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MACHINE    DESIGN 


MY 


ALBERT    W.   SMJTH 

Director  of  Sibley  College,   Cornell  University 


GUIDO    H.    MARX 

Associate  Professor  of  Mechanical  Engineering 
Leland  Stanford  Junior   L  'niversity 


SECOND  EDITION,  REVISED  AND  ENLARGED 

FIRST  THOUSAND 


NEW  YORK 

JOHN    WILEY    &    SONS 

LONDON:     CHAPMAN   &    HALL,    LIMITED 

1908. 


Copyright,  1905,  1908 

BY 

ALBERT  W.  SMITH  AND  GUIDO   H.   MARX 


ii'hr  fcrirntifir  ^TB 
Enlirrt  Onimimiuii 


PREFACE   TO   THE   SECOND   EDITION. 


ONE  can  never  become  a  machine  designer  by  studying  books. 
Much  help  may  come  from  books,  but  the  true  designer  must 
have  judgment,  ripened  by  experience,  in  constructing  and 
operating  machines.  One  may  know  the  laws  that  govern  the 
development,  transmission  and  application  of  energy;  may  have 
knowledge  of  constructive  materials;  may  know  how  to  obtain 
results  by  mathematical  processes,  and  yet  be  unable  to  design 
a  good  machine.  There  is  also  needed  a  knowledge  of  many 
things  connected  with  manufacture,  transportation,  erection  and 
operation.  With  this  knowledge  it  is  possible  to  take  results 
of  computation  and  accept,  reject  and  modify  until  a  machine 
is  produced  that  will  do  the  required  work  satisfactorily. 

Professor  John  E.  Sweet  once  said,  "It  is  comparatively  easy 
to  design  a  good  new  machine,  but  it  is  very  hard  to  design  a 
machine  that  will  be  good  when  it  is  old."  A  machine  must 
not  only  do  its  work  at  first,  but  must  continue  to  do  it  with  a 
minimum  of  repairs  as  long  as  the  work  needs  to  be  done.  The 
designer  must  be  able  to  foresee  the  results  of  machine  operation; 
he  must  have  imagination.  This  is  an  inborn  power,  but  it  may 
be  developed  by  use  and  by  engineering  experience. 

But  there  is  a  certain  part  of  the  designer's  mental  equipment 
that  may  be  furnished  in  the  class-room,  or  by  books.  This  is 
the  excuse  for  the  following  pages.  Machine  design  cannot  be 
treated  exhaustively.  There  are  too  many  kinds  of  machines 
for  this  and  their  differences  are  too  great.  In  this  book  an 

iii 


iv  PREFACE. 

effort  is  made  simply  to  give  principles  that  underlie  all  machine 
design  and  to  suggest  methods  of  reasoning  which  may  be  helpful 
in  the  designing  of  any  machine.  A  knowledge  of  the  usual 
university  course  in  pure  and  applied  mathematics  is  pre- 
supposed. 


INTRODUCTION. 


IN  general  there  are  four  considerations  of  prime  importance 
in  designing  machines:  I.  Adaptation,  II.  Strength  and  Stiff- 
ness, III.  Economy,  IV.  Appearance. 

I.  This  requires  all  complexity  to  be  reduced  to  its  lowest 
terms  in  order  that  the  machine  shall  accomplish  the  desired 
result  in  the  most  direct  way  possible,  and  with  greatest  convenience 
to  the  operator. 

II.  This  requires  the  machine  parts  subjected  to  the  action  of 
forces  to  sustain  these  forces,  not  only  without  rupture,  but  also 
without  such  yielding  as  would  interfere  with  the  accurate  action 
of  the  machine.     In  many  cases  the  forces  to  be  resisted  may 
be  calculated,  and  the  laws  of  Mechanics  and  the  known  qualities 
of  constructive  materials  become  factors  in  determining  propor- 
tions.    In  other  cases  the  force,  by  the  use  of  a  "breaking-piece," 
may  be  limited  to  a  maximum  value,  which  therefore  dictates 
the  design.     But  in  many  other  cases  the  forces  acting  are  neces- 
sarily unknown ;   and  appeal  must  be  made  to  the  precedent  of 
successful  practice,  or  to  the  judgment  of  some  experienced  man, 
until  one's  own  judgment  becomes  trustworthy  by  experience. 

In  proportioning  machine  parts,  the  designer  must  always  be 
sure  that  the  stress  which  is  the  basis  of  the  calculation  or  the 
estimate,  is  the  maximum  possible  stress;  otherwise  the  part 
will  be  incorrectly  proportioned.  For  instance,  if  the  arms  of  a 
pulley  were  to  be  designed  solely  on  the  assumption  that  they 


vi  INTRODUCTION. 

endure  only  the  transverse  stress  due  to  the  belt  tension,  they 
would  be  found  to  be  absurdly  small,  because  the  stresses  resulting 
from  the  shrinkage  of  the  casting  in  cooling  are  often  far  greater 
than  those  due  to  the  belt  pull. 

The  design  of  many  machines  is  a  result  of  what  may  be  called 
"machine  evolution."  The  first  machine  was  built  according  to 
the  best  judgment  of  its  designer;  but  that  judgment  was  fallible, 
and  some  part  yielded  under  the  stresses  sustained;  it  was  replaced 
by  a  new  part  made  stronger;  it  yielded  again,  and  again  was 
enlarged,  or  perhaps  made  of  some  more  suitable  material;  it 
then  sustained  the  applied  stresses  satisfactorily.  Some  other 
part  yielded  too  much  under  stress,  although  it  was  entirely  safe 
from  actual  rupture;  this  part  was  then  stiffened  and  the  process 
continued  till  the  whole  machine  became  properly  proportioned 
for  the  resisting  of  stress.  Many  valuable  lessons  have  been  learned 
from  this  process;  many  excellent  machines  have  resulted  from 
it.  There  are,  however,  two  objections  to  it:  it  is  slow  and  very 
expensive,  and  if  any  part  had  originally  an  excess  of  material, 
it  is  not  changed;  only  the  parts  that  yield  are  perfected. 

Modern  analytical  methods  are  rightly  displacing  it  in  all 
progressive  establishments. 

III.  The  attainment  of  economy  does  not  necessarily  mean  the 
saving  of  metal  or  labor,  although  it  may  mean  that.  To  illustrate : 
Suppose  that  it  is  required  to  design  an  engine-lathe  for  the 
market.  The  competition  is  sharp;  the  profits  are  small.  How 
shall  the  designer  change  the  design  of  the  lathes  on  the  market 
to  increase  profits  ?  (a)  He  may,  if  possible,  reduce  the  weight 
of  metal  used,  maintaining  strength  and  stiffness  by  better  dis- 
tribution. But  this  must  not  increase  labor  in  the  foundry  or 
machine-shop,  nor  reduce  weight  which  prevents  undue  vibrations. 
(6)  He  may  design  special  tools  to  reduce  labor  without  reduction 
of  the  standard  of  workmanship.  The  interest  on  the  first  cost 
of  these  special  tools,  however,  must  not  exceed  the  possible  gain 


INTRODUCTION.  vii 

from  increased  profits.  (c)  He  may  make  th^  lathe  more  con- 
venient for  the  workmen.  True  economy  permits  some  increase 
in  cost  to  gain  this  end.  It  is  not  meant  that  elaborate  and 
expensive  devices  are  to  be  used,  such  as  often  come  from  men 
of  more  inventiveness  than  judgment;  but  that  if  the  parts  can 
be  rearranged,  or  in  any  way  changed  so  that  the  lathes-man 
shall  select  this  lathe  to  use  because  it  is  handier,  when  other 
lathes  are  available,  then  economy  has  been  served,  even  though 
the  cost  has  been  somewhat  increased,  because  the  favorable 
opinion  of  intelligent  workmen  means  increased  sales. 

In  (a)  economy  is  served  by  a  reduction  of  metal;  in  (6)  by  a 
reduction  of  labor;  in  (c)  it  may  be  served  by  an  increase  of  both 
labor  and  material. 

The  addition  of  material  largely  in  excess  of  that  necessary 
for  strength  and  rigidity,  to  reduce  vibrations,  may  also  be  in  the 
interest  of  economy,  because  it  may  increase  the  durability  of  the 
machine  and  its  foundation;  may  reduce  the  expense  incident 
upon  repairs  and  delays,  thereby  bettering  the  reputation  of  the 
machine  and  increasing  sales. 

Suppose,  to  illustrate  further,  that  a  machine  part  is  to  be 
designed,  and  either  of  two  forms,  A  or  B,  will  serve  equally  well. 
The  part  is  to  be  of  cast  iron.  The  pattern  for  A  will  cost  twice 
as  much  as  for  B.  In  the  foundry  and  machine-shop,  however, 
A  can  be  produced  a  very  little  cheaper  than  B-  Clearly  then  -if 
but  one  machine  is  to  be  built,  B  should  be  decided  on ;  whereas, 
if  the  machine  is  to  be  manufactured  in  large  numbers,  A  is 
preferable.  Expense  for  patterns  is  a  first  cost.  Expense  for 
work  in  the  foundry  and  machine-shop  is  repeated  with  each 
machine. 

Economy  of  operation  also  needs  attention.  This  depends 
upon  the  efficiency  of  the  machine ;  i.e.,  upon  the  proportion  of  the 
energy  supplied  to  the  machine  which  really  does  useful  work. 
This  efficiency  is  increased  by  the  reduction  of  useless  frictional 


viii  INTRODUCTION. 

resistances,  by  careful  attention  to  the  design  and  means  of  lubri-" 
cation  of  rubbing  surfaces. 

In  order  that  economy  may  be  best  attained,  the  machine 
designer  needs  to  be  familiar  with  all  the  processes  used  in  the 
construction  of  machines — pattern -making,  foundry  work,  forging, 
and  the  processes  of  the  machine-shop — and  must  have  them  con- 
stantly in  mind,  so  that  while  each  part  designed  is  made  strong 
enough  and  stiff  enough,  and  properly  and  conveniently  arranged, 
and  of  such  form  as  to  be  satisfactory  in  appearance,  it  also  is 
so  designed  that  the  cost  of  construction  is  a  minimum. 

IV.  The  fourth  important  consideration  is  Appearance. 
There  is  a  beauty  possible  of  attainment  in  the  design  of  machines 
which  is  always  the  outgrowth  of  a  purpose.  Otherwise  expressed, 
a  machine  to  be  beautiful  must  be  purposeful.  Ornament  for 
ornament's  sake  is  seldom  admissible  in  machine  design.  And 
yet  the  striving  for  a  pleasing  effect  is  as  much  a  part  of  the  duty 
of  a  machine  designer  as  it  is  a  part  of  the  duty  of  an  architect. 

As-  a  guiding  principle,  the  general  rule  may  be  laid  down 
that  simplicity  and  directness  are  always  best.  Each  member 
should  be  studied  with  strict  reference  to  the  function  which  it 
is  to  perform  and  the  stresses  to  which  it  is  subjected  and  then 
given  the  form  and  size  best  suited  to  meet  the  conditions  with 
the  greatest  economy  of  material  and  workmanship.  When 
combined,  the  parts  must  be  modified  in  such  manner  as  may  be 
found  necessary  to  the  harmonious  effect  of  the  whole. 


CONTENTS. 


CHAPTER  I. 

PAGE 

PRELIMINARY  i 

CHAPTER  II. 
MOTION  IN  MECHANISMS 14 

CHAPTER  III. 
PARALLEL  OR  STRAIGHT-LINE  MOTIONS 39 

CHAPTER  IV. 
CAMS ..-.-.. 47 

CHAPTER  V. 
ENERGY  IN  MACHINES  54 

CHAPTER  VI. 
PROPORTIONS  OF  MACHINE  PARTS  AS  DICTATED  BY  STRESS 71 

m 

CHAPTER  VII. 
RIVETED  JOINTS 87 

CHAPTER  VIII. 
BOLTS  AND  SCREWS 119 

CHAPTER  IX. 

MEANS  FOR  PREVENTING  RELATIVE  ROTATION 148 

ix 


X  CONTENTS. 

CHAPTER   X. 

VACK 

SLIDING  SURFACES  l^2 

CHAPTER   XI. 
AXLES,  SHAFTS,  AND  SPINDLES 170 

CHAPTER   XII. 
JOURNALS,  BEARINGS,  AND  LUBRICATION 181 

CHAPTER  XIII. 
ROLLER-  AND  BALL-BEARINGS 210 

CHAPTER  XIV. 
COUPLINGS  AND  CLUTCHES 221 

CHAPTER  XV. 
BELTS   230 

CHAPTER  XVI. 
FLY-WHEELS • 256 

CHAPTER  XVII. 
TOOTHED  WHEELS  OR  GEARS 275 

CHAPTER  XVIII. 
SPRINGS  327 

CHAPTER  XIX. 
MACHINE  SUPPORTS 332 

CHAPTER   XX. 
MACHINE   FRAMES    •  • 337 

APPENDIX 373 

INDEX   381 


MACHINE   DESIGN. 


CHAPTER  I. 

PRELIMINARY. 

i.  Definitions. — The  study  of  machine  design  is  based  upon 
the  science  of  mechanics,  which  treats  questions  involving  the 
consideration  of  motion,  force,  work,  and  energy.  Since  it  will 
be  necessary  to  use  these  terms  almost  continually,  it  is  well  to 
make  an  exact  statement  of  what  is  to  be  understood  by  them. 

Motion  may  be  denned  as  change  of  position  in  space. 

A  Force  is  one  of  a  pair  of  equal,  opposite,  and  simultaneous 
actions  between  two  bodies  by  which  the  state  of  their  motion 
is  altered,  or  a  change  in  the  form  or  condition  of  the  bodies  them- 
selves is  effected. 

Work  is  the  name  given  to  the  result  of  a  force  in  motion. 

Energy  is  the  capacity  possessed  by  matter  to  do  work. 

The  law  of  Conservation  of  Energy  underlies  every  machine 
problem.  This  law  may  be  expressed  as  follows:  The  sum  of 
energy  in  the  universe  is  constant.  Energy  may  be  transferred 
in  space;  it  may  be  stored  for  varying  lengths  of  time;  it  may 
be  changed  from  one  of  its  several  forms  to  another;  but  it  can- 
not be  created  or  destroyed. 

The  application  of  this  law  to  machines  is  as  follows:  A 
machine  receives  energy  from  a  source,  and  uses  it  to  do  useful 
and  useless  work. 


2  MACHINE  DESIGN. 

A  complete  cycle  of  action  of  a  machine  is  such  an  interval 
that  all  conditions  in  the  machine  are  the  same  at  its  beginning 
and  end,  each  member  of  the  machine  having  in  the  mean  time 
gone  through  all  motions  possible  to  it. 

During  a  complete  cycle  of  action  of  the  machine,  the  energy 
received  equals  the  total  work  done.  The  work  done  may  appear 
as  (a)  useful  work  delivered  by  the  machine,  or  as  (b)  heat  due 
to  energy  transformed  through  frictional  resistance,  or  as  (c) 
stored  mechanical  energy  in  some  moving  part  of  the  machine 
whose  velocity  is  increased.  The  sign  of  the  stored  energy  may 
be  plus  or  minus,  so  that  energy  received  in  one  cycle  may  be 
delivered  during  another  cycle;  but  for  any  considerable  time 
interval  of  machine  action  the  algebraic  sum  of  the  stored  energy 
must  equal  zero. 

For  a  single  cycle: 
Energy  received  =  useful  work  +  useless  work  ± stored  energy. 

For  continuous  action: 

Energy  received  =  useful  work  +  useless  work. 
In  operation  a  machine  generally  acts  by  a  continuous  repetition 
of  its  cycle. 

2.  Efficiency  of  Machines. — In   general,    efficiency   may   be 
denned  as  the  ratio  of  a  result  to  the  effort  made  to  produce  that 
result.     In  a  machine  the  result  corresponds  to  the  useful  work, 
while  the  effort  corresponds  to  the  energy  received.     Hence  the 
efficiency  of  a  machine  =  useful  work  -4- energy  received.*    The 
designer   must   strive   for  high  efficiency,  i.e.,   for   the   greatest 
possible  result  for  a  given  effort. 

3.  Function    of    Machines.  —  Nature    furnishes    sources    of 
energy,  and  the  supplying  of  human  needs  requires  work  to  be 
done.    The  function  of  machines  is  to  cause  matter  possessing 
energy  to  do  useful  work. 

*  The  work  and  energy  must,  of  course,  be  expressed  in  the  same  units. 


PRELIMINARY.  3 

The  chief  sources  of  energy  in  nature  available  for  machine 
purposes  are: 

ist.  The  energy  of  air  in  motion  (i.e.,  wind)  due  to  its  mass 
and  velocity. 

2d.  The  energy  of  water  due  to  its  mass  and  motion  or  posi- 
tion. 

3d.  The  energy  dormant  in  fuels  which  manifests  itself  as 
heat  upon  combustion. 

The  general  method  by  which  the  machine  function  is  exer- 
cised may  be  shown  by  the  following  illustration: 

Illustration. — The  water  in  a  mill-pond  possesses  energy 
(potential)  by  virtue  of  its  position.  The  earth  exerts  an  attrac- 
tive force  upon  it.  If  there  is  no  outlet,  the  earth's  attractive 
force  cannot  cause  motion;  and  hence,  since  motion  is  a  neces- 
sary factor  of  work,  no  work  is  done. 

If  the  water  overflows  the  dam,  the  earth's  attraction  causes 
that  part  of  it  which  overflows  to  move  to  a  lower  level,  and  before 
it  can  be  brought  to  rest  again  it  does  work  against  the  force 
which  brings  it  to  rest.  If  this  water  simply  falls  upon  rocks, 
its  energy  is  transformed  into  heat,  with  no  useful  result. 

But  if  the  water  is  led  from  the  pond  to  a  lower  level,  in  a 
closed  pipe  which  connects  with  a  water-wheel,  it  will  act  upon 
the  vanes  of  the  wheel  (because  of  the  earth's  attraction),  and 
will  cause  the  wheel  and  its  shaft  to  rotate  against  resistance, 
whereby  it  may  do  useful  work.  The  water-wheel  is  a  machine 
and  is  called  a  Prime  Mover,  because  it  is  the  first  link  in  the 
machine-chain  between  natural  energy  and  useful  work. 

Since  it  is  usually  necessary  to  do  the  required  work  at  som.3 
distance  from  the  necessary  location  of  the  water-wheel,  Machinery 
of  Transmission  is  used  (shafts,  pulleys,  belts,  cables,  etc.),  and 
the  rotative  energy  is  rendered  available  at  the  required  place. 

But  this  rotative  energy  may  not  be  suitable  to  do  the  re- 
quired work;  the  rotation  may  be  too  slow  or  too  fast;  a  resist- 


4  MACHINE  DESIGN. 

ance  may  need  to  be  overcome  in  straight,  parallel  lines,  or  at 
periodical  intervals.  Hence  Machinery  of  Application  is  intro- 
duced to  transform  the  energy  to  meet  the  requirements  of  the 
work  to  be  done.  Thus  the  chain  is  complete,  and  the  potential 
energy  of  the  water  does  the  required  useful  work. 

The  chain  of  machines  which  has  the  steam-boiler  and  engine 
for  its  prime  mover  transforms  the  potential  heat  energy  of 
fuel  into  useful  work.  This  might  be  analyzed  in  a  similar  way. 

4.  Free  Motion. — The  general  science  of  mechanics  treats  of 
the  action  of  forces  upon  "free  bodies." 

In  the  case  of  a  "  free  body  "  acted  on  by  a  system  of  forces 
not  in  equilibrium,  motion  results  in  the  direction  of  the  resultant 
of  the  system.  If  another  force  is  introduced  whose  line  of 
action  does  not  coincide  with  that  of  the  resultant,  the  line  of 
action  of  the  resultant  is  changed,  and  the  body  moves  in  a  new 
direction.  The  character  of  the  motion,  therefore,  is  dependent 
upon  the  forces  which  produce  the  motion.  This  is  called  free 
motion. 

Example. — In  Fig.  i,  suppose  the  free  body  M  to  be  acted 
on  by  the  concurrent  forces  i,  2,  and  3 
whose  lines  of  action  pass  through  the 
center  of  gravity  of  M.  The  line  of 
action  of  the  resultant  of  these  forces  is 
A  Bj  and  the  body's  center  of  gravity  would 
move  along  this  line. 

If  another  force,  4,  is  introduced,  CD 
FrG-  T-  becomes  the  line  of  action  of  the  resultant, 

and  the  motion  of  the  body  is  along  the  line  CD. 

5.  Constrained     Motion. — In     a     machine     certain     definite 
motions  occur;    any  departure  from  these  motions,  or  the  pro- 
duction of   any  other    motions,  would    result    in    derangement 
of  the  action  of  the  machine.    Thus,  the  spindle  of  an  engine- 
lathe   turns  accurately  about  its  axis;    the   cutting-tool  moves 


PRELIMINARY.  5 

parallel  to  the  spindle's  axis;  and  an  accurate  cylindrical  surface 
is  thereby  produced.  If  there  were  any  departure  from  these 
motions,  the  lathe  would  fail  to  do  its  required  work.  In  all 
machines  certain  definite  motions  must  be  produced,  and  all 
other  motions  must  be  prevented;  or,  in  other  words,  motion 
in  machines  must  be  constrained. 

Constrained  motion  differs  from  free  motion  in  being  inde- 
pendent of  the  forces  which  produce  it.  If  any  force,  not  suffi- 
ciently great  to  produce  deformation,  be  applied  to  a  body  whose 
motion  is  constrained,  the  result  is  either  a  certain  predeter- 
mined motion,  or  no  motion  at  all. 

6.  Force  Opposed  by  Passive  Resistance. — A  force  may  act 
without  being  able  to  produce  motion  (and  hence  without  being 
able  to  do  work),  as  in  the  case  of  the  water  in  a  mill-pond  without 
overflow  or  outlet.  This  may  be  further  illustrated:  Suppose  a 
force,  say  hand  pressure,  to  be  applied  vertically  to  the  top  of  a 
table.  The  material  of  the  table  offers  a  passive  resistance,  and 
the  force  is  unable  to  produce  motion-,  or  to  do  work. 

It  is  therefore  possible  to  offer  passive  resistance  to  such 
forces  as  may  be  required  not  to  produce  motion,  thereby  render- 
ing them  incapable  of  doing  work.  Whenever  a  body  opposes 
a  passive  resistance  to  the  action  of  a  force  a  change  in  its  condi- 
tion is  effected:  the  force  sets  up  an  equivalent  stress  ia  the 
material  of  the  body.  Thus,  when  the  table  offers  a  passive 
resistance  to  the  hand -pressure,  compressive  stress  is  induced 
in  the  legs.  In  every  case  the  material  of  the  body  must  be  of 
such  shape  and  strength  as  to  resist  successfully  the  induced 
stress. 

In  a  machine  there  must  be  provision  for  resisting  every 
possible  force  which  tends  to  produce  any  but  the  required  motion. 
This  provision  is  usually  made  by  means  of  the  passive  resistance 
of  properly  formed  and  sufficiently  resistant  metallic  surfaces. 

Illustration  I. — Fig.  2  represents  a  section   and  end  view  of 


6  MACHINE  DESIGN. 

a  wood-lathe  headstock.     It  is  required  that  the  spindle,  S,  an  1 
the  attached  cone  pulley,  C,  shall  have  no  other  motion  than 


FIG.  2. 

rotation  about  the  axis  of  the  spindle.  If  any  other  motion  is 
possible,  this  machine  part  cannot  be  used  for  the  required  pur- 
pose. At  A  and  B  the  cylindrical  surfaces  of  the  spindle  are 
enclosed  by  accurately  fitted  bearings  or  internal  cylindrical  sur- 
faces. Suppose  any  force,  P,  whose  line  of  action  lies  in  the 
plane  of  the  paper,  to  be  applied  to  the  cone  pulley.  It  may  be 
resolved  into  a  radial  component,  R,  and  a  tangential  component, 
T.  The  passive  resistance  of  the  cylindrical  surfaces  of  the 
journal  and  its  bearing,  prevents  R  from  producing  motion; 
while  it  offers  no  resistance,  friction  being  disregarded,  to  the 
action  of  T,  which  is  allowed  to  produce  the  required  motion, 
i.e.,  rotation  about  the  spindle's  axis.  If  the  line  of  action  of  P 
pass  through  the  axis,  its  tangential  component  becomes  zero, 
and  no  motion  results.  If  the  line  of  action  of  P  become  tangen- 
tial, its  radial  component  becomes  zero,  and  P  is  wholly  applied 
to  produce  rotation.  If  a  force  Q,  whose  line  of  action  lies  in 
the  plane  of  the  paper,  be  applied  to  the  cone,  it  may  be  resolved 
into  a  radial  component,  N,  and  a  component,  M,  parallel  to 
the  spindle's  axis.  N  is  resisted  as  before  by  the  journal  and 
bearing  surfaces,  and  M  is  resisted  by  the  shoulder  surfaces  of 
the  bearings,  which  fit  against  the  shoulder  surfaces  of  the  cone 
pulley.  The  force  Q  can  therefore  produce  no  motion  at  all. 


PRELIMINARY.  ^ 

In  general,  any  force  applied  to  the  cone  pulley  may  be 
resolved  into  a  radial,  a  tangential,  and  an  axial  component. 
Of  these  only  the  tangential  component  is  able  to  produce  motion; 
and  that  motion  is  the  motion  required.  The  constrainment  is 
therefore  complete;  i.e.,  there  can  be  no  motion  except  rotation 
about  the  spindle's  axis.  This  result  is  due  to  the  passive  resist- 
ance of  metallic  surfaces. 

Illustration  II. — R,  Fig.  3,  represents,  with  all  details  omitted, 
the  "ram,"  or  portion  of  a  shaping-machine  which  carries  the 


FIG.  3. 

cutting-tool.  It  is  required  to  produce  plane  surfaces,  and  hence 
the  "ram"  must  have  accurate  rectilinear  motion  in  the  direction 
of  HK.  Any  deviation  from  such  motion  would  render  the 
machine  useless. 

Consider  Fig.  3,  A.  Any  force  which  can  be  applied  to  the 
ram  may  be  resolved  into  three  components:  one  vertical,  one 
horizontal  and  parallel  to  the  paper,  and  one  perpendicular  to 
the  paper.  The  vertical  component,  if  acting  upward,  is  resisted 
by  the  plane  surfaces  in  contact  at  C  and  D;  if  acting  downward, 
it  is  resisted  by  the  plane  surfaces  in  contact  at  E.  Therefore 
no  vertical  component  can  produce  motion.  The  horizontal 
component  parallel  to  the  paper  is  resisted  by  the  plane  surfaces 
in  contact  at  F  or  G,  according  as  it  acts  toward  the  right  or 
left.  The  component  perpendicular  to  the  paper  is  free  to  pro- 
duce motion  in  the  direction  of  its  line  of  action ;  but  this  is  the 
motion  required. 

Any  force,  therefore,  which  has  a  component  perpendicular 
to  the  paper  can  produce  the  required  motion,  but  no  other 


8  MACHINE  DESIGN. 

motion.  The  constrainment  is  therefore  complete,  and  the 
result  is  due  to  the  passive  resistance  offered  by  metallic  surfaces. 

Complete  Constrainment  is  not  always  required  in  machines. 
It  is  only  necessary  to  prevent  such  motions  as  interfere  with 
the  accomplishment  of  the  desired  result. 

The  weight  of  a  moving  part  is  sometimes  utilized  to  produce 
constrainment  in  one  direction.  Thus  in  a  planer-table,  and  in 
some  lathe-carriages,  downward  motion  and  unallowable  side 
motion  are  resisted  by  metallic  surfaces;  while  upward  motion 
is  resisted  by  the  weight  of  the  moving  part. 

From  the  foregoing  it  follows  that,  as  passive  resistances 
can  be  opposed  to  all  forces  whose  lines  of  action  do  not  coincide 
with  the  desired  direction  of  motion  of  any  machine  part,  it  may 
be  said  that  the  nature  of  the  motion  is  independent  of  the  forces 
producing  it. 

Since  the  motions  0}  machine  parts  are  independent  of  the  jorces 
producing  them,  it  follows  that  the  relation  of  such  motions  may 
be  determined  without  bringing  force  into  the  consideration. 

7.  Kinds  of  Motion  in  Machines. — Motion  in  machines  may 
be  very  complex,  but  it  is  chiefly  plane  motion. 

When  a  body  moves  in  such  a  way  that  any  section  of  it  re- 
mains in  the  same  plane,  its  motion  is  called  plane  motion.  All 
sections  parallel  to  the  above  section  must  also  remain,  each  in 
its  own  plane.  If  the  plane  motion  is  such  that  all  points  of  the 
moving  body  remain  at  a  constant  distance  from  some  line,  AB, 
the  motion  is  called  rotation  about  the  axis  AB.  Example. — 
A  line-shaft  with  attached  parts. 

If  all  points  of  a  body  move  in  straight  parallel  paths,  the 
motion  of  the  body  is  called  rectilinear  translation.  Examples. — 
Engine  cross-head,  lathe-carriage,  planer-table,  shaper-ram. 
Rectilinear  translation  may  be  conveniently  considered  as  a 
special  case  of  rotation,  in  which  the  axis  of  rotation  is  at  an 
infinite  distance,  at  right  angles  to  the  motion. 

If  a  body  moves  parallel  to  an  axis  about  which  it  rotates, 


PRELIMINARY. 


the  body  is  said  to  have  helical  or  screw  motion.    Example. — 
A  nut  turning  upon  a  stationary  screw. 

If  all  points  of  a  body,  whose  motion  is  not  plane  motion, 
move  so  that  their  distances  from  a  certain  point,  O,  remain 
constant,  the  motion  is  called  spheric  motion.  This  is  because 
each  point  moves  in  the  surface  of  a  sphere  whose  center  is  O. 
Example. — The  arms  of  a  fly-ball  steam-engine  governor,  when 
the  vertical  position  is  changing. 

8.  Relative  Motion. — The  motion  of  any  machine  part,  like 
all  known  motion,  is  relative  motion.  It  is  studied  by  reference 
to  some  other  part  of  the  same  machine.  Some  one  part  of  a 
machine  is  usually  (though  not  necessarily)  fixed,  i.e.,  it  has  no 
motion  relative  to  the  earth.  This  fixed  part  is  called  the  frame 
of  the  machine.  The  motion  of  a  machine  part  may  be  referred 
to  the  frame,  or,  as  is  often  necessary,  to  some  other  part  which 
also  has  motion  relative  to  the  frame. 

The  kind  and  amount  of  relative  motion  of  a  machine  part 
depend  upon  the  motions  of  the  part  to  which  its  motion  is 
referred. 

Illustration. — Fig.  4  shows  a  press.  A  is  the  frame;  C  is  a 
plate  which  is  so  constrained  that,  its 
motion  being  referred  to  A,  it  may 
move  vertically,  but  cannot  rotate. 
Motion  oi  rotation  is  communicated  to 
the  screw  B.  The  motion  of  B  re- 
ferred to  A  is  helical  motion,  i.e., 
combined  rotation  and  translation. 
C,  however,  shares  the  translation  of 
B,  and  hence  there  is  left  only  rotation 
as  the  relative  motion  of  B  and  C.  FIG.  4. 

The  motion  of  B  referred  to  C  is  rotation.     The  motion  of  C  re- 
ferred to  B  is  rotation.  The  motion  of  C  referred  to  A  is  translation. 

In  general,  if  two  machine  members,  M  and  N,  move  relative 
to  a  third  member,  /?,  the  relative  motion  of  M  referred  to  N 


TO  MACHINE  DESIGN. 

depends  on  how  much  of  the  motion  of  N  is  shared  by  M.  If 
M  and  N  have  the  same  motions  relative  to  R,  they  have  no  mo- 
tion relative  to  each  other. 

Conversely,  if  two  bodies  have  no  relative  motion,  they  have 
the  same  motion  relative  to  a  third  body.  Thus  in  Fig.  4,  if 
the  constrainment  of  C  were  such  that  it  could  share  5's  rotation, 
as  well  as  its  translation,  then  C  would  have  helical  motion  rela- 
tive to  the  frame,  and  no  motion  at  all  relative  to  B.  This  is 
assumed  to  be  self-evident. 

A  rigid  body  is  one  in  which  the  distance  between  elementary 
portions  *  is  constant.  No  body  is  absolutely  rigid,  but  usually 
in  machine  members  the  departure  from  rigidity  is  so  slight  that 
it  may  be  neglected. 

Many  machine  members,  as  springs,  etc.,  are  useful  because 
of  their  lack  of  rigidity. 

Points  *  in  a  rigid  body  can  have  no  relative  motion,  and  hence 
must  all  have  the  same  motion. 

9.  Instantaneous  Motion  and  Instantaneous  Centers  or  Cen- 
tros. — Points  of  a  moving  body  trace  more  or  less  complex  paths. 
If  a  point  be  considered  as  moving  from 
one  position  in  its  path  to  another  in- 
definitely near,  its  motion  is  called  in- 
stantaneous motion.  The  point  is  mov- 
ing, for  the  instant,  along  a  straight 
line  joining  the  two  indefinitely  near 
together  positions,  and  such  a  line  is 
a  tangent  to  the  path.  In  problems  which  are  solved  by  the 
aid  of  the  conception  of  instantaneous  motion  it  is  only  neces- 
sary to  know  the  direction  of  motion;  hence,  for  such  purposes, 
the  instantaneous  motion  of  a  point  is  fully  defined  by  a  tangent  to 
its  path  through  the  point. 

*  In  this  volume  these  terms  are  used  as  interchangeable  with  the  term 
"particles"  of  mechanics. 


PRELIMINARY.  1 1 

Thus  in  Fig.  5,  if  a  point  is  moving  in  the  path  APB,  when 
it  occupies  the  position  P  the  tangent  TT  represents  its  instan- 
taneous motion.  Any  number  of  curves  could  be  drawn  tangent 
to  TT  at  P,  and  any  one  of  them  would  be  a  possible  path  of 
the  point;  but  whatever  path  it  is  following,  its  instantaneous 
motion  is  represented  by  TT.  The  instantaneous  motion  of  a 
point  is  therefore  independent  of  the  form  of  its  path.  Any  one 
of  the  possible  paths  may  be  considered  as  equivalent,  for  the 
instant,  to  a  circle  whose  center  is  anywhere  in  the  normal  NN. 

In  general,  the  instantaneous  motion  of  a  point,  P,  is  equiva- 
lent to  rotation  about  some  point,  O,  in  a  line  through  the  point  P 
perpendicular  to  the  direction  of  its  instantaneous  motion. 

Let  the  instantaneous  motion  of  a  point,  A,  Fig.  6,  in  a  sec- 
tion of  a  moving  body  be  given  by  the  line  TT.  Then  the  motion 
is  equivalent  to  rotation  about  some  point  on  the  line  AB  as  a 
center,  but  it  may  be  any  point,  and  hence  the  instantaneous 
motion  of  the  body  is  not  determined.  But  if  the  instantaneous 
motion  of  another  point,  C,  be  given  by  the  line  TiTi,  this  motion 
is  equivalent  to  rotation  about  some  point  of  CD.  But  the  points 
A  and  C  are  points  in  a  rigid  body,  and  can  have  no  relative 
motion,  and  must  have  the  same  motion,  i.e.,  rotation  about  the 
same  center.  A  rotates  about  some  point  of  AB,  and  C  rotates 
about  some  point  of  CD;  but  they  must  rotate  about  the  same 
point,  and  the  only  point  which 
is  at  the  same  time  in  both  lines 
is  their  intersection,  O.  Hence 
A  and  C,  and  all  other  points 
of  the  body,  rotate,  for  the  instant, 
about  an  axis  of  which  O  is  the 
projection;  or,  in  other  words,  the 
instantaneous  motion  of  the  body  FlG  6 

is  rotation   about   an  axis  of  which 
O  is    the    projection.     This    axis    is    the   instantaneous   axis  of 


12  MACHINE  DESIGN. 

the    body's    motion,  and  O  is  the  instantaneous  center   of   the 
motion  of  the  section  shown  in  Fig.  6. 

For  the  sake  of  brevity  an  instantaneous  center  will  be  called 
a  centre. 

If  TT  and  TiTi  had  been  parallel  to  each  other,  AB  and 
CD  would  also  have  been  parallel,  and  would  have  intersected 
at  infinity;  in  which  case  the  body's  instantaneous  motion  would 
have  been  rotation  about  an  axis  infinitely  distant;  i.e.,  it  would 
have  been  translation. 

The  motion  of  the  body  in  Fig.  6  is  of  course  referred  to  a 
fixed  body,  which,  in  this  case,  may  be  represented  by  the  paper. 
The  instantaneous  motion  of  the  body  relative  to  the  paper  is 
rotation  about  O.  Let  M  represent  the  figure,  and  N  the  fixed 
body  represented  by  the  paper.  Suppose  the  material  of  M 
to  be  extended  so  as  to  include  O.  Then  a  pin  could  be  put 
through  O,  materially  connecting  M  and  N,  without  interfering 
with  their  instantaneous  motion.  Such  connection  at  any  other 
point  would  interfere  with  the  instantaneous  motion. 

The  centra  of  the  relative  motion  0}  two  bodies  is  a  point,  and 
the  only  one,  at  which  they  have  no  relative  motion;  it  is  a  point, 
and  the  only  one,  that  is  common  to  the  two  bodies  for  the  instant. 

It  will  be  seen  that  the  points  of  the  figure  in  Fig.  6  might 
be  moving  in  any  paths,  so  long  as  those  paths  are  tangent  at 
the  points  to  the  lines  representing  the  instantaneous  motion. 

In  general,  centres  of  the  relative  motion  of  two  bodies  are 
continually  changing  their  position.  They  may,  however,  remain 
stationary;  i.e.,  they  may  become  fixed  centers  of  rotation. 

10.  Loci  of  Centres,  or  Centrodes.* — As  centros  change  posi- 
tion they  describe  curves  of  some  kind,  and  these  loci  of  centros 
may  be  called  cenlrodes. 

*  Centrode  is  here  used  in  preference  to  "centroid,"  proposed  by  Professor 
Kennedy,  because  the  latter  term  has  grown  to  be  generally  accepted  in  mathe- 
matics as  synonymous  with  "center  of  mass." 


PRELIMINARY.  13 

Suppose  a  section  of  any  body,  M,  to  have  motion  relatively 
to  a  section  of  another  body,  N  (fixed),  in  the  same  or  a  parallel 
plane.  Centres  may  be  found  for  a  series  of  positions,  and  a 
curve  drawn  through  them  on  the  plane  of  N  would  be  the  cen- 
trode  of  the  motion  of  M  relatively  to  N.  If,  now,  M  being 
fixed,  N  moves  so  that  the  relative  motion  is  the  same  as  before, 
the  centrode  of  the  motion  of  N  relatively  to  M  may  be  located 
upon  the  plane  of  M.  Now,  since  the  centra  of  the  relative 
motion  of  two  bodies  is  a  point  at  which  they  have  no  relative 
motion,  and  since  the  points  of  the  centrodes  become  succes- 
sively the  centres  of  the  relative  motion,  it  follows  that  as  the 
motion  goes  on,  the  centrodes  would  roll  upon  each  other  without 
slipping.  Therefore,  if  the  centrodes  are  drawn,  and  rolled 
upon  each  other  without  slipping,  the  bodies  M  and  N  will  have 
the  same  relative  motion  as  before.  From  this  it  follows  that 
the  relative  plane  motion  of  two  bodies  may  be  reproduced  by 
rolling  together,  without  slipping,  the  centrodes  of  that  motion. 

ii.  Pairs  of  Motion  Elements. — The  external  and  internal 
surfaces  by  which  motion  is  constrained,  as  in  Figs.  2  and  3,  may 
be  called  pairs  of  motion  elements.  The  pair  in  Fig.  2  is  called 
a  turning  pair,  and  the  pair  in  Fig.  3  is  called  a  sliding  pair. 

The  helical  surfaces  by  which  a  nut  and  screw  engage  with 
each  other  are  called  a  twisting  pair.  These  three  pairs  of 
motion  elements  have  their  surfaces  in  contact  throughout.  They 
are  called  lower  pairs.  Another  class,  called  higher  pairs,  have 
contact  only  along  elements  of  their  surfaces.  Examples. — Cams 
and  toothed  wheels. 


CHAPTER   II. 

MOTION   IN  MECHANISMS. 

12.  Linkages  or  Motion  Chains;  Mechanisms. 

In  Fig.  7,  b  is  joined  to  c  by  a  turning  pair; 
c          "          d     "   sliding     " 
d          "          a      li   turning    " 
a          "          b     " 


FIG.  7. 

Evidently  there  is  complete  constrainment  of  the  relative 
motion  of  a,  b,  c,  and  d.  For,  d  being  fixed,  if  any  motion  occurs 
in  either  a,  b,  or  c,  the  other  two  must  have  a  predetermined 
corresponding  motion. 

c  may  represent  the  cross-head,  b  the  connecting-rod,  and  a 
the  crank  of  a  steam-engine  of  the  ordinary  type.  If  c  were 
rigidly  attached  to  a  piston  upon  which  the  expansive  force  of 
steam  acts  toward  the  right,  a  must  rotate  about  ad.  This 
represents  a  machine.  The  members  a,  b,  c,  and  d  may  be  repre- 
sented for  the  study  of  relative  motions  by  the  diagram,  Fig.  8. 

This  assemblage  of  bodies,  connected  so  that  there  is  complete 
constrainment  of  motion,  may  be  called  a  motion  chain  or  linkage, 
and  the  connected  bodies  may  be  called  links.  The  chain  shown 
is  a  simple  chain,  because  no  link  is  joined  to  more  than  two 
others.  If  any  links  of  a  chain  are  joined  to  more  than  two 

H 


MOTION  IN  MECHANISMS.  15 

others,  the  chain  is  a  compound  chain.     Examples  will  be  given 
later. 

When  one  link  of  a  chain  is  fixed,  i.e.,  when  it  becomes  the 
standard  to  which  the  motion  of  the  others  is  referred,  the  chain 
is  called  a  mechanism.  Fixing  different  links  of  a  chain  gives 
different  mechanisms.  Thus  in  Fig.  8,  if  d  is  fixed,  the  mechanism 
is  that  which  is  used  in  the  usual  type  of  steam-engine,  as  in 
Fig.  7.  It  is  called  the  slider-crank  mechanism. 

But  if  a  is  fixed,  the  result  is  an  entirely  different  mechanism ; 
for  b  would  then  rotate  about  the  permanent  center  ab,  d  would 
rotate  about  the  permanent  center  ad,  while  c  would  have  a  more 
complex  motion,  rotating  about  a  constantly  changing  centro, 
whose  path  may  be  found. 


FIG.  8. 

Fixing  b  or  c  would  give,  in  each  case,  a  still  different  mechan- 
ism. 

13.  Location  of  Centres. — In  Fig.  8  d  is  fixed  and  it  is  re- 
quired to  find  the  centers  of  rotation,  either  permanent  or  in- 
stantaneous, of  the  other  three  links.  The  motion  of  a,  relative 
to  the  fixed  link  d,  is  rotation  about  the  fixed  center  ad.  The 
motion  of  c  relative  to  d  is  translation,  or  rotation  about  a  centro 


1 6  MACHINE  DESIGN. 

cd,  at  infinity  vertically.  The  link  b  has  a  point  in  common 
with  a;  it  is  the  centre,  ab,  of  their  relative  motion.  This  point 
may  be  considered  as  a  point  in  a  or  b;  in  either  case  it  can  have 
but  one  direction  of  motion  relative  to  any  one  standard.  As  a 
point  in  a  its  motion,  relative  to  d,  is  rotation  about  ad.  For  the 
instant,  then,  it  is  moving  along  a  tangent  to  the  circle  through  ab. 
But,  as  a  point  in  b,  its  direction  of  instantaneous  motion  relative 
to  d  must  be  the  same,  and  hence  its  motion  must  be  rotation 
about  some  point  in  the  line  ad-ab,  extended  if  necessary.  Also, 
b  has  a  point,  be,  in  common  with  c;  and  by  the  same  reasoning 
as  above,  be,  as  a  point  in  b,  rotates  for  the  instant  about  some 
point  of  the  vertical  line  through  be.  Now  ab  and  be  are  points 
of  a  rigid  body,  and  one  rotates  for  the  instant  about  some  point 
of  AB,  and  the  other  rotates  for  the  instant  about  some  point 
of  CD;  hence  both  ab  and  be  (as  well  as  all  other  points  of  b) 
must  rotate  about  the  intersection  of  AB  and  CD.  Hence  bd 
is  the  centre  of  the  motion  of  b  relative  to  d. 

The  motion  of  a  may  be  referred  to  c  (fixed),  and  ac  will  be 
found  (by  reasoning  like  that  applied  to  6)  to  lie  at  the  inter- 
section of  the  lines  EF  and  GH. 

The  motion  chain  in  Fig.  8,  as  before  stated,  is  called  the 
slider-crank  chain. 

14.  Centres  of  the  Relative  Motion  of  Three  Bodies  are  always 
in  the  Same  Straight  Line.- — In  Fig.  8  it  will  be  seen  that  the 
three  centros  of  any  three  links  lie  in  the  same  straight  line. 
Thus  ad,  ab,  and  bd  are  the  centros  of  the  links  a,  b,  and  d.  This 
is  true  of  any  other  set  of  three  links. 

Proof. — Consider  a,  b,  and  d.  The  centre  ab  as  a  point  in  a 
has  a  direction  of  instantaneous  motion  relative  to  d  perpen- 
dicular to  a  line  joining  it  to  ad.  As  a  point  in  b  it  has  a  direc- 
tion of  instantaneous  motion  relative  to  d  perpendicular  to  a  line 
joining  it  to  bd.  Therefore  the  lines  ab-ad  and  ab-bd  are  both 
perpendicular  to  the  direction  of  instantaneous  motion  of  ab,  and 


MOTION  IN  MECHANISMS.  17 

they  also  both  pass  through  ab;  hence  they  must  coincide,  and 
therefore  ab,  ad,  and  bd  must  lie  in  the  same  straight  line.  But  a, 
b,  and  d  might  be  any  three  bodies  whatever  which  have  relative 
plane  motion,  and  the  above  reasoning  would  hold.  Hence  it  may 
be  stated :  The  three  centres  of  any  three  bodies-  having  relative 
plane  motion  must  lie  in  the  same  straight  line.  (The  statement 
and  proof  of  this  important  proposition  is  due  to  Prof.  Kennedy.) 
15.  Lever-crank  Chain.  Location  of  Centres. — Fig.  9  shows 
a  chain  of  four  links  of  unequal  length  joined  to  each  other  by 


FIG.  9. 

turning  pairs.  The  centres  ab,  ad,  cd,  and  be  may  be  located  at 
once,  since  they  are  at  the  centers  of  turning  pairs  which  join 
adjacent  links  to  each  other.  The  centres  of  the  relative  motion 
of  b,  c,  and  d  are  be,  cd,  and  bd,  and  these  must  be  in  the  same 
straight  line.  Hence  bd  is  in  the  line  B.  The  centres  of  the 
relative  motion  of  a,  b,  and  d  are  ab,  bd,  and  ad',  and  these  also 
must  lie  in  a  straight  line.  Hence  bd  is  in  the  line  A.  Being 
at  the  same  time  in  A  and  B,  it  must  be  at  their  intersection. 
By  employing  the  same  method  ac  may  be  found. 

16.  The  Constrainment  of  Motion  in  a  linkage  is  independent 
of  the  size  of  the  motion  elements.  As  long  as  the  cylindrical 
surfaces  of  turning  pairs  have  their  axes  unchanged,  the  surfaces 
themselves  may  .be  of  any  size  whatever,  and  the  motion  is  un- 
changed. The  same  is  true  of  sliding  and  twisting  pairs. 


1 8  MACHINE  DESIGN. 

In  Fig.  10,  suppose  the  turning  pair  connecting  c  and  d  to  be 
enlarged  so  that  it  includes  be.     The  link  c  now  becomes  a 


ab 


FIG.  10. 

cylinder,  turning  in  a  ring  attached  to,  and  forming  part  of,  the 
link  d.  be  becomes  a  pin  made  fast  in  c  and  engaging  with  an  eye 
at  the  end  of  b.  The  centres  are  the  same  as  before  the  enlarge- 
ment of  the  pair  cd,  and  hence  the  relative  motion  is  the  same. 

In  Fig.  ii  the  circular  portion  immediately  surrounding  cd 
is  attached  to  d.  The  link  c  now  becomes  a  ring  moving  in  a 
circular  slot.  This  may  be  simplified  as  in  Fig.  12,  whence  c 
becomes  a  curved  block  moving  in  a  limited  circular  slot  in  d. 
The  centres  remain  as  before,  the  relative  motion  is  the  same, 
and  the  linkage  is  essentially  unchanged. 

If,  in  the  slider-crank  mechanism,  the  turning  pair  whose 
axis  is  ab  be  enlarged  till  ad  is  included,  as  in  Fig.  13,  the  motion 
of  the  mechanism  is  unchanged,  but  the  link  a  is  now  called 
an  eccentric  instead  of  a  crank.  This  mechanism  is  usually 
used  to  communicate  motion  from  the  main  shaft  of  a  steam-engine 
to  the  valve.  It  is  used  because  it  may  be  put  on  the  main  shaft 
anywhere  without  interfering  with  its  continuity  and  strength. 

17.  Slotted  Cross-head. — The  mechanism  shown  in  Fig.  14 
is  called  the  " slotted  cross-head  mechanism."  Its  centres  may 
be  found  from  principles  already  given. 


MOTION  IN  MECHANISMS.  19 

This  mechanism  is  often  used  as  follows:  One  end  of  c,  as 
E,  is  attached  to  a  piston  working  in  ^  cylinder  attached  to  d. 
This  piston  is  caused  to  reciprocate  by  the  expansive  force  of 
steam  or  some  other  fluid.  The  other  end  of  c  is  attached  to 


FIG.  ii. 


ab 


FIG.  12. 

another  piston,  which  also  works  in  a  cylinder  attached  to  d. 
This  piston  may  pump  water  or  compress  gas  (for  example 
small  ammonia  compressors  for  refrigerating  plants).  The 
crank  a  is  attached  to  a.  shaft,  the  projection  of  whose  axis  is 
ad.  This  shaft  also  carries  a  fly-wheel  which  insures  approxi- 
mately uniform  rotation. 


MACHINE  DESIGN. 


18.  Location  of  Centros  in  a  Compound  Mechanism. — It  is 
required  to  find  the  centres  of  the  compound  linkage,  Fig.  15. 
In  any  linkage,  each  link  has  a  centre  relatively  to  every  other 


FIG.  13. 

link;   hence,  if  the  number  of  links  =  n,  the  number  of  centros  = 
n(n  —  i).     But  the  centro  ab  is  the  same  as  ba;  i.e.,  each  centra 


FIG.  14. 
is  double.    Hence  the  number  of  centros  to  be  located  for  any 


linkage  =  ' 


.     In  the  linkage  Fig.  15,  the  number  of  centros 


*  The  links  are   a,  b,  c,  d,  e,  and  /. 
The  centros:     ab  be  cd  de  ef 
ac  bd  ce   df 
ad  be   c} 
ae  bf 
a) 


MOTION  IN  MECHANISMS. 


21 


The  portion  above  the  link  d  is  a  slider-crank  chain,  and 
the  character  of  its  motion  is  in  no  way  affected  by  the  attachment 
of  the  part  below  d.  On  the  other  hand,  the  lower  part  is  a 
lever-crank  chain,  and  the  character  of  its  motion  is  not  affected 
by  its  attachment  to  the  upper  part.  The  chain  may  therefore 
be  treated  in  two  parts,  and  the  centres  of  each  part  may  be 
located  from  what  has  preceded.  Each  part  will  have  six  centros, 
and  twelve  would  thus  be  located,  ad,  however,  is  common  to 


FIG.  15. 

the  two  parts,  and  hence  only  eleven  are  really  found.  Four 
centros,  therefore,  remain  to  be  located.  They  are  be,  c],  b],  and 
ce.  To  locate  be,  consider  the  three  links  a,  b,  and  e,  and  it 
follows  that  be  is  in  the  line  A  passing  through  ab  and  ae;  con- 
sidering b,  d,  and  e,  it  follows  that  be  is  in  the  line  B  through  bd 
and  de.  Hence  be  is  at  the  intersection  of  A  and  B.  Similar 
methods  locate  the  other  centros. 

In  general,  for  finding  the  centros  of  a  compound  linkage  of 


22  MACHINE  DESIGN. 

six  links,  consider  the  linkage  to  be  made  up  of  two  simple  chains, 
and  find  their  centres  independently  of  each  other.  Then  take 
the  two  links  whose  centre  is  required,  together  with  one  of 
the  links  carrying  three  motion  elements  (as  a,  Fig.  15).  The 
centres  of  these  links  locate  a  straight  line,  A,  which  contains 
the  required  centra.  Then  take  the  two  links  whose  centra  is 
required,  together  with  the  other  link  which  carries  three  motion 
elements.  A  straight  line,  B,  is  thereby  located,  which  contains 
the  required  centra,  and  the  latter  is  therefore  at  the  intersection 
of  A  and  B. 

19.  Velocity  is  the  rate  of  motion,  or  motion  per  unit  time. 

Linear  velocity  is  linear  space  moved  through  in  unit  time; 
it  may  be  expressed  in  any  units  of  length  and  time;  as,  miles 
per  hour,  feet  per  minute  or  per  second,  etc. 

Angular  velocity  is  angular  space  moved  through  in  unit  time. 
In  machines,  angular  velocity  is  usually  expressed  in  revolutions 
per  minute  or  per  second. 

The  linear  space  described  by  a  point  in  a  rotating  body,  or 
its  linear  velocity,  is  directly  proportional  to  its  radius,  or  its 
distance  from  the  axis  of  rotation.  This  is  true  because  arcs 
are  proportional  to  radii. 

If  A  and  B  are  two  points  in  a  rotating  body,  and  if  r\  and  r% 
are  their  radii,  then  the  ratio  of  linear  velocities 

_  linear  veloc.  A  _  r\_ 
~  linear  veloc.  B    r2' 

This  is  true  whether  the  rotation  is  about  a  center  or  a  centra; 
i.e.,  it  is  true  either  for  continuous  or  instantaneous  rotation. 
Hence  it  applies  to  all  cases  of  plane  motion  in  machines;  because 
all  plane  motion  in  machines  is  equivalent  to  either  continuous 
or  instantaneous  rotation  about  some  point. 

To  find  the  relation  of  linear  velocity  of  two  points  in  a  machine 
member,  therefore,  it  is  only  necessary  to  find  the  relation  of 


MOTION  IN  MECHANISMS.  23 

the  radii  of  the  points.  The  latter  relation  can  easily  be  found 
when  the  center  or  centre  is  located. 

20.  A  vector  quantity  possesses  magnitude  and  direction.  It 
may  be  represented  by  a  straight  line,  because  the  latter  has 
magnitude  (its  length)  and  direction.  Thus  the  length  of  a 
straight  line,  AB,  may  represent,  upon  some  scale,  the  magnitude 
of  some  vector  quantity,  and  it  may  represent  the  vector  quantity's 
direction  by  being  parallel  to  it,  or  by  being  perpendicular  to  it. 
For  convenience  the  latter  plan  will  here  be  used.  The  vector 
quantities  to  be  represented  are  the  linear  velocities  of  points 
in  mechanisms.  The  lines  which  represent  vector  quantities  are 
called  vectors. 

A  line  which  represents  the  linear  velocity  of  a  point  will 
be  called  the  linear  velocity  vector  of  the  point.  The  symbol  of 
linear  velocity  will  be  VI.  Thus  VIA  is  the  linear  velocity  of 
the  point  A.  Also  Va  will  be  used  as  the  symbol  of  angular 
velocity. 

If  the  linear  velocity  and  radius  of  a  point  are  known,  the 
angular  velocity,  or  the  number  of  revolutions  per  unit  time, 
may  be  found;  since  the  linear  velocity -r- length  of  the  circum- 
ference in  which  the  point  travels  =  angular  velocity. 

All  points  of  a  rigid  body  have  the  same  angular  velocity. 

If  the  radii,  and  ratio  of  linear  velocities  of  two  points,  in 
different  machine  members  are  known,  the  ratio  of  the  angular 
velocities  of  the  members  may  be  found  as  follows: 

Let  A  be  a  point  in  a  member  M,  and  B  a  point  in  a  member 
N.  TI  =  radius  of  A ;  r2  =  radius  of  B.  VIA  and  VIB  represent 

VIA 
the  linear  velocities  of  A    and  B,   whose   ratio,  T™,  is  known. 

Then  VaA=—    and     VaB=— . 

27iri  27ZT2 

VaA_VlA     27TT2_VIA     r2     VaM 
Hence  -         X         -         X     - 


24  MACHINE  DESIGN. 

If  M  and  N  rotate  uniformly  about  fixed  centers,  the  ratio 
v  jy.  is  constant.  If  either  M  or  N  rotates  about  a  centre,  the 

ratio  is  a  varying  one. 

21.  To  find  the  relation  of  linear  velocity  of  two  points  in 
the  same  link,  it  is  only  necessary  to  measure  the  radii  of  the 
points,  and  the  ratio  of  these  radii  is  the  ratio  of  the  linear  veloci- 
ties of  the  points. 

In  Fig.  1 6,  let  the  smaller  circle  represent  the  path  of  Ar 
the  center  of  the  crank-pin  of  a  slider-crank  mechanism;  the 
link  d  being  fixed.  Let  the  larger  circle  represent  the  rim  of  a 
pulley  which  is  keyed  to  the  same  shaft  as  the  crank.  The 
pulley  and  the  crank  are  then  parts  of  the  same  link.  The  ratio 

VIA 
of  velocity  of  the  crank-pin  center  and  the  pulley  surf  ace =~^TJJ> 

=— .     In  this  case  the  link  rotates  about  a  fixed  center.     The 
f\ 

same  relation  holds,   however,   when  the  link  rotates  about  a 
centro. 


FIG.  1 6. 
22.  Velocity  Diagram  of  Slider-crank  Chain. — In    Fig.    17, 

the  link  d  is  fixed  and  -TTJT-  =  ,  _ ,  . .     By  similar  triangles  this 
expression  is  also  equal  to    n_A  .     Hence,  if  the  radius  of  the 


MOTION  IN  MECHANISMS.  25 

crank  circle  be  taken  as  the  vector  o]  the  constant  linear  velocity 
of  ab,  the  distance  cut  off  on  the  vertical  through  O  by  the  line  oj 
the  connecting-rod  (extended  i]  necessary)  will  be  the  vector  of  the 
linear  velocity  0}  be.  Project  A  horizontally  upon  bc-bd,  locating 
B.  Then  bc-B  is  the  vector  of  VI  of  the  slider,  and  may  be 

FIG.  17. 


FIG.  18. 

used  as  an  ordinate  of  the  linear  velocity  diagram  of  the  slider. 
By  repeating  the  above  construction  for  a  series  of  positions, 
the  ordinates  representing  the  VI  of  be  for  different  positions  of 
the  slider  may  be  found.  A  smooth  curve  through  the  extremi- 
ties of  these  ordinates  is  the  velocity  curve,  from  which  the  Vis 


26  MACHINE  DESIGN. 

of  all  points  of  the  slider's  stroke  may  be  read.  The  scale  of 
velocities,  or  the  linear  velocity  represented  by  one  inch  of  ordi- 
nate,  equals  the  constant  linear  velocity  of  ab  divided  by  O-ab 
in  inches. 

23.  Velocity  Diagram  of  Lever-crank  Chain. — It  is  required 
to  find  VI  of  be  during  a  cycle  of  action  of  the  mechanism  shown 
in  Fig.  1 8,  d  being  fixed,  and  VI  of  ab  being  constant.  The 
two  points  ab  and  be  may  both  be  considered  in  the  link  b. 
All  points  in  b  move  about  bd  relatively  to  the  fixed  link. 

Vlab     ab-bd 

Hence  T77,    =-7 — ry. 

Vibe      bc-bd 

For  most  positions  of  the  mechanism  bd  will  be  so  located  as  to 
make  it  practically  impossible  to  measure  these  radii,  but  a  line, 
as  MN,  drawn  parallel  to  b  cuts  off  on  the  radii  portions  which 
are  proportional  to  the  radii  themselves,  and  hence  proportional 
to  the  Vis  of  the  points.  Hence 

Vlab     ab-M 


Vibe      bc-N ' 

The  arc  in  which  be  moves  may  be  divided  into  any  number  of 
parts,  and  the  corresponding  positions  of  ab  may  be  located.  A 
circle  through  M,  with  ad  as  center,  may  be  drawn,  and  the 
constant  radial  distance  ab-M  may  represent  the  constant 
velocity  of  ab.  Through  Mi,  If 2,  etc.,  draw  lines  parallel  to  the 
corresponding  positions  of  b,  and  these  lines  will  cut  off  on  the 
corresponding  line  of  c  a  distance  which  represents  VI  of  be. 
Through  the  points  thus  determined  the  velocity  diagram  may 
be  drawn,  and  the  VI  of  be  for  a  complete  cycle  is  determined. 
The  scale  of  velocities  is  found  as  in  Sec.  22. 

24.  The  relation  of  linear  velocity  of  points  not  in  the  same 
link  may  also  be  found. 


MOTION  IN  MECHANISMS. 


27 


VI  of  A 


Required      .    ^ '     referred   to  d  as  the  fixed  link,   Fig.    19. 

The  centro  ab  is  a  point  in  common  to  a  and  b,  the  two  links 
considered.  Consider  ab  as  a  point  in  a;  and  its  VI  is  to  that 
of  A  as  their  radii  or  distances  from  ad.  Draw  a  vector  triangle 
with  its  sides  parallel  to  the  triangle  formed  by  joining  A,  ab, 


FIG.  19. 

and  ad.  Then  if  the  side  A\  represent  the  VI  of  A,  the  side  aib\ 
will  represent  the  VI  of  ab.  Consider  ab  as  a  point  in  b,  and 
its  VI  is  to  that  of  B  as  their  radii,  or  distances  to  bd.  Upon 
the  vector  atbi  draw  a  triangle  whose  sides  are  parallel  to  those 
of  a  triangle  formed  by  joining  ab,  bd,  and  B.  Then,  from 
similar  triangles,  the  side  B\  is  the  vector  of  .B's  linear  velocity. 


Hence 


VI  of  A     vector  A  i 
VI  of  5  =  vector  B\' 


The  path  of  B  during  a  complete  cycle  may  be  traced,  and  the 
VI  for  a  series  of  points  may  be  found,  by  the  above  method,  then 
the  vectors  may  be  laid  off  on  normals  to  the  path  through  the 
points;  the  velocity  curve  may  be  drawn;  and  the  velocity  of 
B  at  all  points  becomes  known. 


23 


MACHINE  DESIGN. 


25.  Angularity  of  Connecting-rod. — The  diagram  of  VI  of 
the  slider-crank  mechanism,  Fig.  17,  is  unsymmetrical  with 
respect  to  a  vertical  axis  through  its  center.  This  is  due  to  the 
angularity  of  the  connecting-rod,  and  may  be  explained  as  follows : 

In  Fig.  20,  AO  is  one  angular  position  of  the  crank,  and  BO 
is  the  corresponding  angular  position  on  the  other  side  of  the 
vertical  through  the  center  of  rotation.  The  corresponding 
positions  of  the  slider  are  as  shown.  But  for  position  A  the  line 
of  the  connecting-rod,  C,  cuts  off  on  the  vertical  through  O  a 
vector  Oa,  which  represents  the  slider's  velocity.  For  position 
B  the  vector  of  the  slider's  velocity  is  Ob  and  the  velocity  diagram 
is  unsymmetrical. 


FIG.  20. 

If  the  connecting-rod  were  parallel  to  the  direction  of  the 
slider's  motion  in  all  positions,  as  in  the  slotted  cross-head 
mechanism  (see  Fig.  14),  the  vector  cut  off  on  the  vertical  through 
O  would  be  the  same  for  position  A  and  position  B  and  the 
velocity  diagram  would  be  symmetrical. 

Since  the  velocity  diagram  is  symmetrical  with  a  parallel 
connecting-rod  and  unsymmetrical  with  an  angular  connecting- 
rod,  with  all  other  conditions  constant,  it  follows  that  the  lack 
of  symmetry  is  due  to  the  angularity  of  the  connecting-rod. 

The  velocity  diagram  for  the  slotted  cross-head  mechanism 
is  symmetrical  with  respect  to  both  vertical  and  horizontal  axes 
through  its  center.  In  fact,  if  the  crank  radius  (  =  length  of 


MOTION  IN  MECHANISMS.  29 

link  a)  be  taken  as  the  vector  of  the  VI  of  ab,  the  linear  velocity 
diagram  of  the  slider  becomes  a  circle  whose  radius  =  the  length  of 
the  link  a.  Hence  the  crank  circle  itself  serves  for  the  linear  velocity 
diagram,  the  horizontal  diameter  representing  the  path  of  the  slider. 
26.  Angularity  of  Connecting-rod,  Continued. — During  a  por- 
tion of  the  cycle  of  the  slider-crank  mechanism,  the  slider's  VI 
is  greater  that  than  of  ab.  This  is  also  due  to  the  angularity  of 
the  connecting-rod,  and  may  be  explained  as  follows:  In  Fig.  21, 
as  the  crank  moves  up  from  the  position  x,  it  will  reach  such  a 
position,  A,  that  the  line  of  the  connecting-rod  extended  will 
pass  through  B.  OB  in  this  position  is  the  vector  of  the  linear 
velocity  of  both  ab  and  the  slider,  and  hence  their  linear  velocities 


FIG.  21. 

are  equal.  When  ab  reaches  B,  the  line  of  the  connecting-rod 
passes  through  B;  and  again  the  vectors — and  hence  the  linear 
velocities — of  ab  and  the  slider  are  equal.  For  all  positions 
between  A  and  B  the  line  of  the  connecting-rod  will  cut  OB 
outside  of  the  crank  circle;  and  hence  the  linear  velocity  of  the 
slider  will  be  greater  than  that  of  ab.  This  result  is  due  to  the 
angularity  of  the  connecting-rod,  because  if  the  latter  remained 
always  horizontal,  its  line  could  never  cut  OB  outside  the  circle. 
It  follows  that  in  the  slotted  cross-head  mechanism  the  maximum 
VI  of  the  slider  =  the  constant  VI  of  ab.  The  angular  space  BOA, 
Fig.  21,  throughout  which  VI  of  the  slider  is  greater  than  the  VI 
of  ab,  increases  with  increase  of  angularity  of  the  connecting-rod; 
i.e.,  it  increases  with  the  ratio 

Length  of  crank 
Length  of  connecting-rod  * 


30  MACHINE  DESIGN. 

27.  Quick-return  Mechanisms.  —  A  slider  in  a  mechanism 
often  carries  a  cutting-tool,  which  cuts  during  its  motion  in  one 
direction,  and  is  idle  during  the  return  stroke.     Sometimes  the 
slider  carries  the  piece  to  be  cut,  and  the  cutting  occurs  while 
it  passes  under  a  tool  made  fast  to  the  fixed  link,  the  return  stroke 
being  idle. 

The  velocity  of  cutting  is  limited.  If  the  limiting  velocity 
be  exceeded,  the  tool  becomes  so  hot  that  its  temper  is  drawn, 
and  it  becomes  unfit  for  cutting.  The  limit  of  cutting  velocity 
depends  on  the  nature  of  the  material  to  be  cut.  Thus  annealed 
tool-steel  and  the  scale  surface  of  cast  iron  may  be  cut  with 
carbon  tool-steel  at  10  to  20  feet  per  minute;  wrought  iron  and 
soft  steel  at  25  to  30  feet  per  minute;  while  brass  and  the  softer 
alloys  may  be  cut  at  40  or  more  feet  per  minute.  With  certain 
special  tool-steels  these  speeds  may  be  considerably  exceeded. 
There  is  no  limit  of  this  kind,  however,  to  the  velocity  during 
the  idle  stroke ;  and  it  is  desirable  to  make  it  as  great  as  possible, 
in  order  to  increase  the  product  of  the  machine.  This  leads 
to  the  design  and  use  of  "quick-return"  mechanisms. 

28.  Slider-crank    Quick    Return.  —  If,    in    a    slider-crank 
mechanism,  the  center  of   rotation  of   the  crank  be  moved,  so 
that  the  line  of  the  slider's  motion  does  not  pass  through  it,  the 
slider  will  have  a  quick-return  motion. 

In  Fig.  22,  when  the  slider  is  in  its  extreme  position  at  the 
right,  A,  the  crank-pin  center  is  at  D.  When  the  slider  is  at  B, 
the  crank-pin  center  is  at  C.  If  rotation  is  as  indicated  by  the 
arrow,  then,  while  the  slider  moves  from  B  to  A,  the  crank-pin 
center  moves  from  C  over  to  D.  And  while  the  slider  returns 
from  A  to  B,  the  crank-pin  center  moves  under  from  D  to  C. 
If  the  VI  of  the  crank-pin  center  be  assumed  constant,  the  time 
occupied  in  moving  from  D  to  C  is  less  than  that  from  C  to  Z). 
Hence  the  time  occupied  by  the  slider  in  moving  from  B  to  A 
is  greater  than  that  occupied  in  moving  from  A  to  B.  The 


MOTION  IN  MECHANISMS.  31 

mean  velocity  during  the  forward  stroke  is  therefore  less  than 
during  the  return  stroke.  Or  the  slider  has  a  "quick-return" 
motion. 


FIG.  22. 

It  is  required  to  design  a  mechanism  of  this  kind  for  a  length 
of  stroke  =BA  and  for  a  ratio 

mean  VI  forward  stroke  _  5 
mean  VI  return  stroke  ™  7  * 

The  mean  velocity  of  either  stroke  is  inversely  proportional  to 
the  time  occupied,  and  the  time  is  proportional  to  the  correspond- 
ing angle  described  by  the  crank.  Hence 

mean  velocity  forward     5     angle  /? 
mean  velocity  return      7      angle  a 

It  is  therefore  necessary  to  divide  360°  into  two  parts  which 
are  to  each  other  as  5  to  7.  Hence  a  =  210°  and  /?  =  i5o°.  Ob- 
viously 0  =  i8o°  —  /?  =  3o°.  Place  the  30°  angle  of  a  drawing 
triangle  so  that  its  sides  pass  through  B  and  A.  This  condition 
may  be  fulfilled  and  yet  the  vertex  of  the  triangle  may  occupy 
an  indefinite  number  of  positions.  By  trial  O  may  be  located  so 
that  the  crank  shall  not  interfere  with  the  line  of  the  slider.* 

*To  avoid  cramping  of  the  mechanism,  the  angle  BAD  should  equal  or  exceed 
i35°- 


32  MACHINE  DESIGN. 

O  being  located  tentatively,  it  is  necessary  to  find  the  correspond- 
ing lengths  of  crank  a  and  connecting-rod  b.  When  the  crank- 
pin  center  is  at  D,  AO  =  b-a;  when  it  is  at  C,  BO=b  +  a.  AO 
and  BO  are  measurable  values  of  length;  hence  a  and  b  may 
be  found,  the  crank  circle  may  be  drawn,  and  the  velocity  dia- 
grams may  be  constructed  as  in  Fig.  17;  remembering  that  the 
distance  cut  off  upon  a  vertical  through  O,  by  the  line  of  the 
connecting-rod,  is  the  vector  of  the  VI  of  the  slider  for  the  corre- 
sponding position  when  the  VI  of  the  crank-pin  center  is  repre- 
sented by  the  crank  radius. 

It  is  required  to  make  the  maximum  velocity  of  the  forward 
stroke  of  the  slider  =  20  feet  per  minute,  and  to  find  the  corre- 
sponding number  of  revolutions  per  minute  of  the  crank.  The 
maximum  linear  velocity  vector  of  the  forward  stroke  =  the 
maximum  height  of  the  upper  part  of  the  velocity  diagram; 
call  it  Vl\.  Call  the  linear  velocity  vector  of  the  crank-pin  center 
F/2=  crank  radius.  Let  x=  linear  velocity  of  the  crank-pin 
center.  Then 

y/i     20  ft.  per  minute 

V12  =         ~^~         ' 

20  ft.  per  minute  X  F/2 

-fir 

x  is  therefore  expressed  in  known  terms.  If  now  x,  the  space 
the  crank-pin  center  is  required  to  move  through  per  minute, 
be  divided  by  the  space  moved  through  per  revolution,  the  result 
will  equal  the  number  of  revolutions  per  minute  =N', 


N  = 


2?rX  length  of  crank* 


29.  Lever-crank  Quick  Return. — Fig.  23  shows  a  compound 
mechanism.  The  link  d  is  the  supporting  frame  or  fixed  link, 
and  a  rotates  about  ad  in  the  direction  indicated,  communicating 


MOTION  IN  MECHANISMS.  33 

motion  to  c  through  the  slider  b  so  that  c  vibrates  about  cd.  The 
link  e,  connected  to  c  by  a  turning  pair  at  ce,  causes  /  to  slide 
horizontally  on  another  part  of  the  frame  or  fixed  link  d.  The 
center  of  the  crank-pin,  ab,  is  given  a  constant  linear  velocity, 
and  the  slider,  /,  has  motion  toward  the  left  with  a  certain  mean 
velocity,  and  returns  toward  the  right  with  a  greater  mean  velocity. 
This  is  true  because  the  slider  /  moves  toward  the  left  while  a 
moves  through  the  angle  a ;  and  toward  the  right  while  a  moves 
through  the  angle  ft.  But  the  motion  of  a  is  uniform,  and  hence 
the  angular  movement  a  represents  more  time  than  the  angular 
movement  /3;  and  /,  therefore,  has  more  time  to  move  toward 
the  left  than  it  has  to  move  through  the  same  space  toward  the 
right.  It  therefore  has  a  "quick-return"  motion. 

/<•/ 

J- />/> 


\       .^1\\J 

md* 


FIG.  23. 

The  machine  is  driven  so  that  the  crank-pin  center  moves 
uniformly,  and  the  velocity,  at  all  points  of  its  stroke,  of  the 
slider  carrying  a  cutting-tool,  is  required.  The  problem,  there- 
fore, is  to  find  the  relation  of  linear  velocities  of  ef  and  ab  for  a 
series,  of  positions  during  the  cycle;  and  to  draw  the  diagram 
of  velocity  of  ef. 

Solution. — ab  has  a  constant  known  linear  velocity.  The 
point  in  the  link  c  which  coincides,  for  the  instant,  with  ab,  re- 
ceives motion  from  ab,  but  the  direction  of  its  motion  is  different 


34  MACHINE  DESIGN. 

from  that  of  ab,  because  ab  rotates  about  ad,  while  the  coin- 
ciding point  of  c  rotates  about  cd.  If  ab-A  be  laid  off  repre- 
senting the  linear  velocity  of  ab,  then  ab-B  will  represent  the 
linear  velocity  of  the  coinciding  point  of  the  link  c.  Let  the 
latter  point  be  called  x. 

Locate  cf,  at  the  intersection  of  e  with  the  line  cd-ad.  Now 
c}  and  x  are  both  points  in  the  link  c,  and  hence  their  linear 
velocities,  relatively  to  the  fixed  link  d,  are  proportional  to  their 
distances  from  cd.  These  two  distances  may  be  measured 
directly,  and  with  the  known  value  of  linear  velocity  of  x  =  ab-B 
give  three  known  values  of  a  simple  proportion,  from  which  the 
fourth  term,  the  linear  velocity  of  cf,  may  be  found. 

Or,  if  the  line  BD  be  drawn  parallel  to  cd-ad,  the  triangle 
B-D-ab  is  similar  to  the  triangle  cd-cj-ab,  and  from  the  simi- 
larity of  these  triangles  it  follows  that  BD  represents  the  linear 
velocity  of  cf  on  the  same  scale  that  ab-B  represents  the  linear 
velocity  of  x.  Hence  the  linear  velocity  of  cf,  for  the  assumed 
position  of  the  mechanism,  becomes  known.  But  since  cf  is  a 
point  of  the  slider,  all  of  whose  points  have  the  same  linear  velocity 
because  its  motion  relatively  to  d  is  rectilinear  translation,  it 
follows  that  the  linear  velocity  of  cf  is  the  required  linear  velocity 
of  the  slider.  At  ef  erect  a  line  perpendicular  to  the  direction 
of  motion  of  the  slider  having  a  length  equal  to  BD. 

This  solution  may  be  made  for  as  many  positions  of  the 
mechanism  as  are  necessary  to  locate  accurately  the  velocity 
curve.  The  ordinates  of  this  curve  will,  of  course,  be  the  veloci- 
ties of  the .  slider,  and  the  abscissae  the  corresponding  positions 
of  the  slider. 

Having  drawn  the  velocity  diagram,  suppose  that  it  is  required 
to  make  the  maximum  linear  velocity  of  the  slider  on  the  slow 
stroke  =Q  feet  per  minute.  Then  the  linear  velocity  of  the 
crank-pin  center  ab=y  can  be  determined  from  the  propor- 
tion 


MOTION  IN  MECHANISMS.  3$ 

y  vector  A-ab 

Q  "maximum  ordinate  of  velocity  diagram' 
vector  A-ab 

•          At   =  fj  . 

7        maximum  ordinate  of  velocity  diagram* 
If  r  =  the  crank  radius,  the  number  of  revolutions  per  minute  =  — . 

When  this  mechanism  is  embodied  in  a  machine,  a  becomes 
a  crank  attached  to  a  shaft  whose  axis  is  at  ad.  The  shaft  turns 
in  bearings  provided  in  the  machine  frame.  The  crank  carries  a 
pin  whose  axis  is  at  ab,  and  this  pin  turns  in  a  bearing  in  the 
sliding  block  b.  The  link  c  becomes  a  lever  keyed  to  a  shaft 
whose  axis  is  at  cd.  This  lever  has  a  long  slot  in  which  the  block 
b  slides.  The  link  e  becomes  a  connecting-rod,  connected  to  both 
c  and  /  by  pin  and  bearing.  The  link  /  becomes  the  "cutter- 
bar"  or  "ram"  of  a  shaper:  the  part  which  carries  the  cutting- 
tool.  The  link  d  becomes  the  frame  of  the  machine,  which  not 
only  affords  support  to  the  shafts  at  ad  and  cd,  and  the  guiding- 
surfaces  for  /,  but  also  is  so  designed  as  to  afford  means  for  holding 
the  pieces  to  be  planed,  and  supports  the  feed  mechanism. 


FIG.  24. 

30.  Whitworth  Quick  Return. — Fig.  24  shows  another  com- 
pound linkage,     d  is  fixed,  and  c  rotates  uniformly  about  cd, 


36  MACHINE  DESIGN. 

communicating  an  irregular  rotary  motion  to  a  through  the  slider 
b.  a  is  extended  past  ad  (the  part  extended  being  in  another 
parallel  plane),  and  moves  a  slider  /  through  the  medium  of 
a  link  e.  This  is  called  the  "Whitworth  quick-return  mechanism." 
The  point  be,  at  which  c  communicates  motion  to  a,  moves  along 
<z,  and  hence  the  radius  (measured  from  ad]  of  the  point  at  which 
•a  receives  a  constant  linear  velocity  varies,  and  the  angular 
velocity  of  a  must  vary  inversely.  Hence  the  angular  velocity 
of  a  is  a  maximum  when  the  radius  is  a  minimum,  i.e.,  when 
a  and  c  are  vertical  downward;  and  the  angular  velocity  of  a 
is  minimum  when  the  radius  is  a  maximum,  i.e.,  when  a  and  c 
are  vertical  upward. 

31.  Problem. — To  design  a  Whitworth  Quick  Return  for  a 
given  ratio, 

mean  VI  of  /  forward 
mean  VI  of  /  returning' 

When  the  center  of  the  crank-pin,  C,  reaches  A,  the  point  D  will 
coincide  with  B,  the  link  c  will  occupy  the  angular  position  cd-B, 
and  the  slider  /  will  be  at  its  extreme  position  toward  the  left. 

When  the  point  C  reaches  F,  the  point  D  will  coincide  with 
E,  the  link  c  will  occupy  the  angular  position  cd-E,  and  the 
slider  /  will  be  at  its  extreme  position  toward  the  right. 

Obviously,  while  the  link  c  moves  over  from  the  position 
cd-E  to  the  position  cd-B,  the  slider  /  will  complete  its  forward 
stroke,  i.e.,  from  right  to  left.  While  c  moves  under  from  cd-B 
to  cd-E,  f  will  complete  the  return  stroke,  i.e.,  from  left  to  right. 
The  link  c  moves  with  a  uniform  angular  velocity,  and  hence  the 
mean  velocity  of  /  forward  is  inversely  proportional  to  the  angle 
/?  (because  the  time  consumed  for  the  stroke  is  proportional  to 
the  angle  moved  through  by  the  crank  c},  and  the  mean  velocity 
of  /  returning  is  inversely  proportional  to  a.  Or 

mean  VI  of  /  forward      a 
mean  VI  of  /  returning    /?' 


MOTION  IN  MECHANISMS.  37 

For  the  design  the  distance  cd-ad  must  be  known.     This  may 
usually  be  decided  on  from  the  limiting  sizes  of  the  journals  at  cd 

and  ad.     Suppose  that  the  above  ratio  =o"  =  y  >  tnat  cd-ad  =3", 

and  that  the  maximum  length  of  stroke  of  /=i2".  Locate  cd 
and  measure  off  vertically  downward  a  distance  equal  to  3", 
thus  locating  ad.  Draw  a  horizontal  line  through  ad.  The 
point  ef  of  the  slider  /  will  move  along  this  line.  Since 

-,      and  a+/?  =  36o0, 


.'.     a  =  150°  and        /?  =  2io°. 

Lay  off  a.  from  cd  as  a  center,  so  that  the  vertical  line  through 
cd  bisects  it.  Draw  a  circle  through  B  with  cd  as  a  center,  5 
being  the  point  of  intersection  of  the  bounding  line  of  a  with  a 
horizontal  through  ad.  The  length  of  the  link  c  =  cd-B. 

The  radius  ad-C  must  equal  the  travel  of  /-*-2  =  6".  This 
radius  is  made  adjustable,  so  that  the  length  of  stroke  may  be 
varied.  The  connecting-rod,  e,  may  be  made  of  any  convenient 
length. 

32.  Problem.  —  To  draw  the  velocity  diagram  of  the  slider 
/  of  the  Whitworth  Quick  Return.  The  point  be,  Fig.  25,  as  a 
point  of  c  has  a  known  constant  linear  velocity  relative  to  d,  and 
its  direction  of  motion  is  always  at  right  angles  to  a  line  joining 
it  to  cd.  That  point  of  the  link  a  which  coincides  in  this  posi- 
tion of  the  mechanism  with  be,  receives  motion  from  be,  but  its 
direction  of  motion  relative  to  d  is  at  right  angles  to  the  line  be- 
ad. If  bc-A  represents  the  linear  velocity  of  be,  its  projection 
upon  bc-ad  extended  will  represent  the  linear  velocity  of  the 
point  of  a  which  coincides  with  be.  Call  this  point  x.  Locate 
the  centre  a/,  draw  the  line  af-bc  and  extend  it  to  meet  the  vertical 
dropped  from  B  to  C.  The  centre  af  may  be  considered  as  a 


3-3 


MACHINE  DESIGN. 


point  in  a,  and  its  linear  velocity  relative  to  d,  when  so  considered, 
is  proportional  to  its  distance  from  ad.     Hence 

VI  oi  af__ad-af 
Vlofx  ~ad-bc' 


FIG.  25. 
But  the  triangles  ad-a}-bc  and  B-C-bc  are  similar.     Hence 

VI  of  af_  BC 
Vlofx~  B-bc 

This  means  that  BC  represents  the  linear  velocity  of  af  upon  the 
same  scale  that  B-bc  represents  the  linear  velocity  of  x.  But 
«/  is  a  point  in  /,  and  all  points  in  /  have  the  same  linear  velocity 
relative  to  d  since  the  motion  is  rectilinear  translation;  hence 
BC  represents  the  linear  velocity  of  the  slider  /  for  the  given 
position  of  the  mechanism,  and  it  may  be  laid  off  as  an  ordinatc* 
of  the  velocity  curve.  This  solution  may  be  made  for  as  many 
positions  as  are  required  to  locate  accurately  the  entire  velocity 
curve  for  a  cycle  of  the  mechanism. 


CHAPTER  III. 

PARALLEL   OR   STRAIGHT-LINE  MOTIONS. 

33.  Watt  Parallel  Motion. — Rectilinear  motion  in  machines 
Is  usually  obtained  by  means  of  prismatic  guides.     It  is  some- 
times  necessary,    however,    to   accomplish    the   same   result   by 
linkages. 

The  simplest  and  most  widely  known  linkage  used  for  giving 
rectilinear  motion  to  a  point  without  the  use  of  any  sliding  pairs 
is  the  so-called  Watt  Parallel  Motion.  It  is  one  of  the  numerous 
inventions  of  James  Watt  and  N 

consists    of    four     links,    three  • 2 ;aj 

moving  and  one  fixed,  all  con- 
nected by  turning  pairs,  d, 
.Fig.  26,  is  the  fixed  link,  a 

rotates  relative  to  d  about  ad; 

i    •  j  j  FlG-  2<5- 

c  rotates  relative  to  d  about  cd. 

The  mechanism  is  shown  in  the  position   corresponding  to  the 
middle  of  its  motion. 

As  the  points  ab  and  be  swing  in  the  dotted  arcs,  the  point  P 
will  travel  in  approximately  a  straight  line.  The  whole  path  of 
P  is  a  lemniscate,  but  the  part  which  is  ordinarily  used  approaches 
very  closely  to  a  straight  line. 

34.  Parallelogram. — A  true  parallel  motion  is  given  by  the 
Parallelogram  which  is  shown  in  Fig.  27.     The  links  a,  b,  c,  and  e 
are  connected  to  each  other  by  turning  pairs,  and  the  linkage  is 
attached  to  the  fixed  link  d  by  a  turning  pair  at  ac.     (This  point 
is  also  ad  and  cd.} 

39 


40  MACHINE  DESIGN. 

The  lengths  ac  —  ab  and  ce—  be  are  equal,  as  are  also  ac—  ce  and 
ab  —  be.  The  point  P  is  fixed  on  e.  Draw  a  line  from  P  to  ac ;  it 
cuts  the  link  &  at  P'. 

By  similar  triangles, 

p'-be      P-be 


ac—ce      P—ce 


.'.  Pr-be 


=  a    constant. 


Therefore  the  point  P'  will  lie  at  the  same  position  on  b  for  all 
positions  of  the  mechanism.     Likewise,  by  similar  triangles,  the 

ratio       ~aC  =  ,  ~Ce    =  a   constant    for   all    positions    of  the 
P'  —  ac       be—ce 

mechanism.     Since  the  line  P—P'  swings,  relative  to  d,  about  the 


ac,ad,cd\    /  \  W 

-,rwi    a 


FIG.  27.  FIG.  28. 

pole  ad  (ac,  cd)  at  every  instant,  it  is  obvious  that  the  motions  of 
P  and  P'  relative  to  d  will  be  similar  to  each  other  in  every  respect 

and  always  in  the  ratio  —     —  •       It  follows  that,  if  either  of 
P'-ac 

these  points  is  guided  to  move  in  a  straight-line  path,  the  other 
point  is  constrained  to  move  in  a  similar  parallel  path.* 

*  The  following  demonstration  is  given  for  those  who  prefer  an  accurate 
proof : 

The  position  of  the  mechanism  in  Fig.  27  is  taken  as  a  perfectly  general  one 
and,  in  the  same  way,  the  instantaneous  motion  of  the  point  P  is  assumed  as 
indicated  by  the  arrow.  By  methods  indicated  in  earlier  chapters  locate  the 
centres  de  and  bd.  Continue  the  lines  ad  —  ab  and  be  —  ab  until  they  cut  the  line 
P-de  at  x  and  y,  respectively.  It  is  necessary  to  prove  that  P'  will  always 


PARALLEL   OR  STRAIGHT-LINE  MOTIONS.  41 

35.  Grasshopper  Motion.  —  A  device  which  may  be  used  to 
change  the  direction  of  rectilinear  motion  through  a  right  angle 
is  the  linkage  known  as  the  Grasshopper  Motion,  shown  in  Fig.  28. 
This  is  the  ordinary  slider-crank  chain  with  crank  a  and  con- 
necting-rod b  of  equal  length.  The  linkage  is  further  modified 

move  in  a  path  parallel  to  P's  motion  and  at  a  constant  proportion  to  it.  If 
this  is  true  for  instantaneous  motion  it  is  true  for  any  motion  P  may  be  given 
relatively  to  d.  Draw  the  line  P' —  bd.  It  can  be  shown  that  this  line  will 
always  be  parallel  to  P  —  de,  for,  since  ac  —  x  is  parallel  to  ce—P 

P-be  _  x-bd 
be—ce     bd—ac' 

p —be      P—Pf 

Also,  by  similar  triangles  -. =  ™ . 

be—ce     P'—ac 

x-bd      P-P' 

Hence,  — =-757 , 

bd—ac     P  —  ac 

which,  considering  the  triangles  ac—x  —  P  and  ac  —  bd—P',  shows  that  P'  —  bd 
is  parallel  to  P—  x,  or  to  P-de. 

But  P  is  a  point  of  e  and  as  such  has  an  instantaneous  motion  relative  to  d  in 
a  direction  perpendicular  to  P  —  de.  In  the  same  way,  P'  is  a  point  of  b  and  as 
such,  relatively  tod,  has  instantaneous  motion  perpendicular  to  P' —  bd.  These 
two  instantaneous  motions  are  therefore  parallel. 

It  remains  to  be  shown  that  they  will  always  be  in  the  same  proportion  as  to 
extent.  The  extent  will  be  directly  proportional  to  the  instantaneous  linear 
velocities.  Both  P  and  be  as  points  of  e  rotate  for  the  instant  about  de 

VIP     P-de 

relatively  to  a,     .".   ^-. — =r — . 

VI  be     be-  de 

Both  be  and  P'  are  points  of  b,  and  as  such,  relatively  to  d,  rotate  about  bd, 
VI  be      b?.-bd      be-de 


'  VIP'     lj'  — 


(by  similar  triangles). 


Multiplying, 

VIP      Vibe      P-de      bf—de  VIP      P—de     , 

X->   °r  =  ~          =  (by  8umlar  tmngles)  ' 


P-be 

j  -  =a  constant  value. 

be—  ce 

Or,  in  other  words,  the  linear  velocities  of  P  and  P'  bear  a  constant  ratio  to  each 
other  for  all  positions  of  the  mechanism,  and  hence,  these  points  will  trace 
proportionately  similar  paths  on  d.  —  Q.E.D. 


MACHINE  DESIGN. 


by  extending  b  beyond  ab  to  a  point  P  such  that  the  length 
P-db  =  crank  length.  It  is  obvious  that  P  is  constrained  to  move 
relative  to  d  in  a  straight-line  path  perpendicular  to  d  through 
ad* 

36.  General  Method  for  Parallel-motion  Design. — A  general 
method  of  design  which  is  applicable  in  many  cases  is  as  follows. 
In  Fig.  29  d  is  the  fixed  link,  and  a  is  connected  with  it  by  a  sliding 
pair,  a,  b,  c,  and  e  are  connected  by  turning  pairs,  as  shown. 
The  constrainment  is  not  complete  because  B  is  free  to  move 
in  any  direction,  and  its  motion  would,  therefore,  depend  upon 
the  force  producing  it.  It  is  required  that  the  point  B  shall  move 


1C 


Fm.  29. 

in  a  straight  line  parallel  to  a.  Suppose  that  B  is  caused  to  move 
along  the  required  line;  then  any  point  of  the  link  c,  as  A,  will 
describe  some  curve,  FAE.  If  a  pin  be  attached  to  c,  with  its 
axis  at  A,  and  a  curved  slot  fitting  the  pin,  with  its  sides  parallel 
to  FAE,  be  attached  to  d,  as  in  Fig.  30,  it  follows  that  B  can  only 
move  in  the  required  straight  line.  This  is  the  mechanism  of  the 
Tabor  Steam-engine  Indicator. 

*  This  is  true  because,  from  the  construction  of  the  mechanism,  the  line  P-bd 
must  always  lie  parallel  to  d.  The  point  Pt  which  rotates  about  bd  as  center  rela- 
tively to  d,  always  has  an  instantaneous  motion  perpendicular  to  P-bd  and,  con- 
•wquently,  perpendicular  to  d. 


PARALLEL    OR  STRAIGHT-LINE  MOTIONS. 


43 


The  curve  described  by  A  might  approximate  a  circular  arc 
whose  center  could  be  located,  say,  at  O,  Fig.  30.    Then  the 


FIG.  30. 

curved  slot  might  be  replaced  by  a  link  attached  to  d  and  c  by 
turning  pairs  at  O  and  A.  This  gives  B  approximately  the 
required  motion.  This  is  the  mechanism  of  the  Thompson 
Steam-engine  Indicator. 

If,  while  the  point  B  is  caused  to  move  in  the  required  straight 
line,  a  point  in  b,  as  P,  Fig.  29,  were  chosen,  it  would  be  found 


FIG.  31. 

to  describe  a  curve  which  would  approximate  a  circular  arc, 
whose  center,  O,  and  radius,  =r,  could  be  found.     Let  the  link 


44 


MACHINE  DESIGN 


whose  length  =  r  be  attached  to  d  and  b  by  turning  pairs  whose 
axes  are  at  O  and  P,  and  the  motion  of  B  will  be  approximately 
the  required  motion.  This  is  the  mechanism  of  the  Crosby 
Steam-engine  Indicator.  One  very  important  fact,  however, 
is  to  be  noted  in  connection  with  all  steam-engine  indicator 
pencil  mechanisms.  While  it  is  important  that  the  pencil  point 


FIG.  3oa. 

B  (Figs.  29  and  30)  travel  in  a  straight-line  path  parallel  to  the 
axis  of  the  piston  rod  a,  it  is  fully  as  important  that  the  motion 
of  the  point  B  always  be  exactly  the  same  multiple  of  a's  motion. 
To  determine  in  any  given  case  whether  this  is  true  or  not,  lay 
off  very  accurately  and  to  a  large  scale,  say  rive  times  actual  size, 
a  skeleton  outline  of  the  mechanism  for  three  positions.  See 
Fig.  300.  These  positions  are  taken  so  that  the  total  distance 


PARALLEL   OR  STRAIGHT-LINE  MOTIONS.  45 

B!~ B3  represents  the  allowable  range  of  the  instrument  as 
stated  by  the  maker,  usually  about  3".  JB2  is  located  at  the  mid- 
position.  The  links  a,  b,  c,  e  and  /are  drawn  for  each  case  in 
their  proper  relative  positions,  d  being  considered  as  the  fixed 
link. 

The  subscripts  i,  2  and  3  refer  to  the  positions  of  the  links 
corresponding  to  the  three  pencil  positions  Bv  B2  and  B3. 

First  take  position  $t.  The  centros  df  and  de  are  located 
at  once  because  they  are  permanent  centers  as  well.  Since  a's 
motion  relative  to  d  is  rectilinear  translation,  the  centro  ad  will 
lie  at  infinity  in  a  direction  at  a  right  angle  to  the  direction  of 
motion,  or,  in  this  case,  at  horizontal  infinity.  The  centros  cfv 
abv  bcv  and  ce1  are  located  at  once  at  the  axes  of  the  turning 
pairs  connecting  the  respective  links. 

Using  Kennedy's  theorem  locate  cd{  (on  lines  de—ce1  and 
df—cfj,  and  acl  (on  lines  cd1—ad  and  ab1  —  bc1}.  At  the 
instant  in  question,  every  point  of  c  relatively  to  d  is  rotating 
about  the  centro  cdl  and  each  point  will  have  a  linear  velocity 
proportional  to  its  distance  from  cdv  But  Bl  and  acl  are  both 

Vlac.       ac.  —  cd 

points  of  c.     Hence  we  may  write          *  =  — ' -1  •       But    ac 

ViB^      Bt  — cdl 

is  also  a  point  of  a  and  at  any  instant  every  point  of  a  has  the 
same  velocity  relatively  to  d  because  the  motion  of  a  relative  to  d 

Via       ac.  —  cd, 

is  rectilinear  translation  .*.  —    -    =  — J *  • 

VlBl      B1—cdi 

Similarly  for  the  second  position, 

VI  a       acn  —  cdn 


And  for  the  third  position, 

Via    _  ac3-  cd, 
VI B3  ~  B3-cd3 


46  MACHINE  DESIGN. 

But,  for  proper  action, 

Via         Via         Via 

=  a  constant  for  all  positions, 

,    should  equal  ^p — V2?    and  also  equal  — ?-', 

Bl-cdl  B2-cd2,'  B3-cd3' 

otherwise  the  diagrams  will  give  a  distortion  of  the  piston,  a's, 
motion.  Also  for  true  parallel  motion  Bl  should  lie  on  the  same 
horizontal  through  cd1  on  which  acl  lies;  B2  on  the  horizontal 
through  cd2;  and  B3  on  the  horizontal  through  cdy 

An  examination  of  existing  indicator  mechanisms  in  this 
manner  gives  very  interesting  results,  and  separates  clearly 
those  instruments  which  distort  from  those  which  are  correct. 

37.  Problem.  —  In  Fig.  31  B  is  the  fixed  axis  of  a  counter- 
shaft; C  is  the  axis  of  another  shaft  which  is  free  to  move  in 
any  direction.  It  is  required  to  constrain  D  to  move  in  the 
straight  line  EF.  If  D  be  moved  along  EF,  a  tracing-point 
fixed  at  A  in  the  link  CD  will  describe  an  approximate  circular 
arc,  HAK,  whose  center  may  be  found  at  O.  A  link  whose 
length  is  OA  may  be  connected  to  the  fixed  link,  and  to  the  link 
CD,  by  means  of  turning  pairs  at  O  and  A.  D  will  then  be 
constrained  to  move  approximately  along  EF.  A  curved  slot 
and  pin  could  be  used,  and  the  motion  would  be  exact.* 

*  Descriptions  of  many  varieties  of  parallel  motions  may  be  found  in  Rankine's 
"Machinery  and  Millwork";  Weisbach's  "Mechanics  of  Engineering,"  Vol.  III^ 
"Mechanics  of  the  Machinery  of  Transmission";  Kennedy's  "Mechanics  of 
Machinery";  and  elsewhere. 


CHAPTER  IV. 

CAMS. 

38.  Cams  Defined. — A  machine  part  of  irregular  outline,  as 
A,  Fig.  32,  may  rotate  or  vibrate  about  an  axis  O,  and  commu- 
nicate motion  by  line  contact  to  another  machine  part,  B.    A  is 
called  a  cam.     A  cylinder  A,  Fig.  33,  having  a  groove  of  any 
form  in  its  surface,  may  rotate  about  its  axis,  CD,  and  communi- 
cate motion  to  another  machine  part,  B.     A  is  a  cam.     A  disk 
A,  Fig.  34,  having  a  groove  in  its  face,  may  rotate  about  its  axis, 
O,   and  communicate  motion  to  another  machine  part,  B.     A 
is  a  cam.     In  fact  it  is  only  a  modification  of  A,  Fig.  32.     In 
designing  cams  it  is  customary  to  consider  a  number  of  simul- 
taneous positions  of  the  driver  and  follower.     The  cam  curve 
can  usually  be  drawn  from  data  thus  obtained. 

39.  Case  I. — The  follower  is  guided  in  a  straight  line,  and 
the  contact  of  the  cam  with  the  follower  is  always  in  this  line. 
The  line  may  be  in  any  position  relatively  to  the  center  of  rota- 
tion of  the  cam;    hence  it  is  a  general  case.     The  point  of  the 
follower  which  bears  on  the  cam  is  constrained  to  move  in  the 
line  MN,  Fig.  35.     O  is  the  center  of  rotation  of  the  cam.     About 
O  as  a  center,  draw  a  circle  tangent  to  MN  at  /.     Then  A, ~B, 
C,  etc.,  are  points  in  the  cam.     When  the  point  A  is  at  J  the 
point  of  the  follower  which  bears  on  the  cam  must  be  at  Af; 
when  B  is  at  /  the  follower- point  must  be  at  Bf;  and  so  on  through 
an  entire  revolution.     Through  A,  B,  C,  etc.,  draw  lines  tangent 

47 


48  MACHINE  DESIGN. 

to  the  circle.  With  O  as  a  center,  and  OA'  as  a  radius,  draw  a 
circular  arc  A'A",  intersecting  the  tangent  through  A  at  A". 
Then  A"  will  be  a  point  in  the  cam  curve.  For,  if  A  returns  to 
/,  AA"  will  coincide  with  JA',  A"  will  coincide  with  A',  and  the 


FIG.  32. 


FIG.  33. 


FIG.  34. 


FIG.  35. 


cam  will  hold  the  follower  in  the  required  position.  The  same 
process  for  the  other  positions  locates  other  points  of  the  cam 
curve.  A  smooth  curve  drawn  through  these  points  is  the 
required  cam  outline.  Often,  to  reduce  friction,  a  roller  attached 
to  the  follower  rests  on  the  cam,  motion  being  communicated 
through  it.  The  curve  found  as  above  will  be  the  path  of  the 
axis  of  the  roller.  The  cam  outline  will  then  be  a  curve  drawn 


CAMS.  49 

inside  of,  and  parallel  to,  the  path  of  the  axis  of  the  roller,  at  a 
•distance  from  it  equal  to  the  roller's  radius.  Contact  between 
the  follower  and  the  cam  is  not  confined  to  the  line  MN  if  a 
roller  is  used. 

40.  Case  II. — The  cam  engages  with  a  surface  of  the  follower, 
and  this  surface  is  guided  so  that  all  of  its  positions  are  parallel. 
The  method  given  is  due  to  Professor  J.  H.  Barr.     O,  Fig.  36, 
is   the   center   of   rotation    of   the   cam.     The   follower   surface 
occupies  the  successive  positions   i,   2,   3,  etc.,  when  the  lines 

A,  B,  C,  etc.,  of  the  cam  coincide  with  the  vertical  line  through 
C.     It  is  required  to  draw  the  outline  of  a  cam   to  produce  the 
motion  required.     Produce  the  vertical  line  through  O,  cutting 
the  positions  of  the  follower  surface  in  A',  Bf,  C',  etc.    With  O 
as  a  center  and  radii  OBf,  OCf,  etc.,  draw  arcs  cutting  the  lines 

B,  C,  D,  etc.,  in  the  points  B",  C",  D" ',  etc.     Position  i  is  the 
lowest  position  of  the  follower  surface;  therefore  A  must  be  in 
contact  with  the  follower  surface  in  the  vertical  line  through  O, 
because  if  the  tangency  be  at  any  other  point  the  motion  in  one 
direction  or  the  other  will  lower  the  follower,  which  is  not  allow- 
able.     A    is   therefore   one   point  in  the   cam   curve.      Draw  a 
line  MN  through  B"  at  right  angles  to  B"O,  and  rotate  B"O 
till  it  coincides  with  B'O.     Then  the  line  MN  will  coincide  with 
the   position   of  the  follower  surface  2Bf.     But   the  cam  curve 
must  be  tangent  to  this  line  when  B  coincides  with  B'O,  and 
therefore  the  line  MN  is  a  line  to  which  the  cam  curve  must  be 
tangent.     Similar  lines  may  be  drawn  through  the  points  C", 
D",  etc.     Each  will  be  a  line  to  which  the  cam  curve  must  be 
tangent.     Therefore,  if  a  smooth  curve  be  drawn  tangent  to  ah 
these  lines,  it  will  be  the  required  cam  outline. 

41.  Case    III. — This  is  the  same  as  Case  II,  except  that  the 
positions  of  the  follower  surface,  instead  of  being  parallel,  con- 
verge to  a  point,  O',  Fig.  37,  about  which  the  follower  vibrates. 
The  solution  is  the  same  as  in  Fig.  36,  except  that  the  angle  be- 


5°  MACHINE  DESIGN. 

tween  the  lines  corresponding  to  MN,  Fig.  36,  and  the  radial 
lines,  instead  of  being  a  right  angle,  equals  the  angle  between 
the  corresponding  position  of  the  follower  surface  and  the  vertical. 


FIG.  36. 


In  these  cases  the  cam  drives  the  follower  in  only  one  direc- 
tion; the  force  of  gravity,  the  expansive  force  of  a  spring,  or 
some  other  force  must  hold  it  in  contact  with  the  cam.  To 
drive  the  follower  in  both  directions,  the  cam  surface  must  be 
double,  i.e.,  it  takes  the  form  of  a  groove  engaging  with  a  pin  or 
roller  attached  to  the  follower,  as  in  Fig.  34. 

This  method  is  inclined  to  produce  excessive  wear.  A  better 
method  is  to  have  the  follower  provided  with  two  rollers  on  op- 
posite sides  of  the  cam-shaft.  See  Fig.  38. 

Cam  A  is  designed  to  give  the  desired  motion  to  the  follower 
through  the  medium  of  roller  i.  Every  position  of  this  roller 
causes  roller  2  to  occupy  a  definite  position,  and  the  complementary 
cam  B  is  so  designed  as  to  correspond  to  these  positions  of  roller 
2.  Cam  B  is  rigidly  mounted  on  the  same  shaft  as  A,  so  that 
the  two  cams  have  no  motion  relative  to  each  other.  If  the  line 
of  action  of  the  follower  passes  through  the  center  of  the  cam- 


CAMS. 


shaft  as  shown  in  Fig.  38,  it  becomes  a  very  simple  matter  to 
draw  the  outline  of  cam  B\   all  that  is  necessary  is  to  keep  the 


Q 


,Q- 


FIG.  38. 

sum  of  the  radial  lengths  a  +  b  =  a  constant  =  the  distance  be- 
tween the  centers  of  rollers  i  and  2. 

42.  Case  IV. — To  lay  out  a  cam  groove  on  the  surface  of  a 
cylinder. — A,  Fig.  39,  is  a  cylinder  which  is  to  rotate  continu- 
ously about  its  axis.  B  can  only  move  parallel  to  the  axis  of  A. 
B  may  have  a  projecting  roller  to  engage  with  a  groove  in  the 
surface  of  A.  CD  is  the  axis  of  the  roller  in  its  mid-position. 
EF  is  the  development  of  the  surface  of  the  cylinder.  During 


52  MACHINE  DESIGN. 

the  first  quarter-revolution  of  A,  CD  is  required  to  move  or.e 
inch  toward  the  right  with  a  constant  velocity.  Lay  off  GH  =  i", 
and  HJ  =  \KF,  locating  /.  Draw  GJ,  which  will  be  the  middle 
line  of  the  cam-groove.  During  the  next  half-revolution  of  A 
the  roller  is  required  to  move  two  inches  toward  the  left  with  a 
uniformly  accelerated  velocity.  Lay  off  JL  =  2",  and  LM  =  %KF. 
Divide  LM  into  any  number  of  equal  parts,  say  four.  Divide 
JL  into  four  parts,  so  that  each  is  greater  than  the  preceding  one 


FIG.  39. 

by  an  equal  increment.  This  may  be  done  as  follows :  1  +  24 
3  +  4  =  10.  Lay  off  from  J,  o.i/L,  locating  a\  then  0.2 JL  from 
a,  locating  b;  and  so  on.  Through  a,  b,  and  c  draw  vertical 
lines;  through  m,  n,  and  o  draw  horizontal  lines.  The  in- 
tersections locate  d,  e,  and  /.  Through  these  points  draw  the 
curve  from  /  to  M,  which  will  be  the  required  middle  line 
of  the  cam-groove.  During  the  remaining  quarter-revolution 
the  roller  is  required  to  return  to  its  starting-point  with  a 


CAMS  53 

uniformly  accelerated  velocity.  The  curve  MN  is  drawn  in 
the  same  way  as  JM.  On  each  side  of  the  line  GJMN  lay  off 
paiallel  lines,  their  distance  apart  being  equal  to  the  diameter  of 
the  roller.  Wrap  EF  upon  the  cylinder,  and  the  required  cam- 
groove  is  located. 


CHAPTER  V. 

ENERGY  IN  MACHINES. 

43.  The  subject  of  motion  and  velocity,  in  certain  simple 
machines,  has  been  treated  and  illustrated.  It  remains  now  to 
consider  the  passage  of  energy  through  similar  machines.  From 
this  the  solution  of  force  problems  will  follow. 

During  the  passage  of  energy  through  a  machine,  or  chain  of 
machines,  any  one,  or  all,  of  four  changes  may  occur. 

I.  The  energy  may  be  transferred  in  space.     Example. — En- 
ergy is  received  at  one  end  of  a  shaft  and  transferred  to  the  other 
end,  where  it  is  received  and  utilized  by  a  machine. 

II.  The  energy  may  be  converted  into  another  form.     Exam- 
ples.— (a)  Heat   energy   into  mechanical  energy   by   the   stea,m- 
engine  machine  chain,     (ft)  Mechanical  energy  into  heat  by  fric- 
tion,     (c)  Mechanical    energy    into  electrical    energy,   as    in   a 
dynamo-electric    machine;    or    electrical  energy  into  mechanical 
energy  in  the  electric  motor,  etc. 

III.  Energy  is  the  product  of  a  force  factor  and  a  space  factor. 
Energy  per  unit  time,  or  rate  oj  doing  work,  is  the  product  of  a 
force  factor  and  a  velocity  factor,  since  velocity  is  space  per  unit 
time.     Either  factor  may  be  changed  at  the  expense  of  the  other; 
i.e.,  velocity  may  be  changed,  if  accompanied  by  such  a  change 
of  force  that  the  energy  per  unit  time  remains  constant.     Cor- 
respondingly, force  may  be  changed  at  the  expense  of  velocity, 
energy  per  unit  time  being  constant.     Example. — A  belt  trans- 
mits 6000  foot-pounds  per  minute  to  a  machine.     The  belt  veloc- 
ity is  120  feet  per  minute,  and  the  force  exerted  is  50  Ibs.     Fric- 

54 


ENERGY  IN  MACHINES.  55 

tional  resistance  is  neglected.  A  cutting-tool  in  the  roachine 
does  useful  work;  its  velocity  is  20  feet  per  minute,  and  the  re- 
sistance to  cutting  is  300  Ibs.  Then,  energy  received  per  minute 
=  120X50  =  6000  foot-pounds;  and  energy  delivered  per  minute 
=  20X300  =  6000  foot-pounds.  The  energy  received  therefore 
equals  the  energy  delivered.  But  the  velocity  and  force  factors 
are  quite  different  in  the  two  cases. 

IV.  Energy  may  be  transferred  in  time.  In  many  machines 
the  energy  received  at  every  instant  equals  that  delivered.  There 
are  many  cases,  however,  where  there  is  a  periodical  demand  for 
work,  i.e.,  a  fluctuation  in  the  rate  of  doing  wcrk;  while  energy 
can  only  be  supplied  at  the  average  rate.  Or  there  may  be  a  uni- 
form rate  of  doing  work,  and  a  fluctuating  rate  of  supplying 
energy.  In  such  cases  means  are  provided  in  the  machine,  or 
chain  of  machines,  for  the  storing  0}  energy  till  it  is  needed.  In 
other  words,  energy  is  transferred  in  time.  Examples. — (a)  In 
the  steam-engine  there  is  a  varying  rate  of  supplying  energy  dur- 
ing each  stroke,  while  there  is  (in  general)  a  uniform  rate  of  doing 
work.  There  is,  therefore,  a  periodical  excess  and  deficiency  of 
effort.  A  heavy  wheel  on  the  main  shaft  absorbs  the  excess  of 
energy  with  increased  velocity,  and  gives  it  out  again  with  re- 
duced velocity  when  the  effort  is  deficient,  (b)  A  pump  delivers 
water  into  a  pipe  system  under  pressure.  The  water  is  used  in  a 
hydraulic  press,  whose  action  is  periodic  and  beyond  the  capacity 
of  the  pump.  A  hydraulic  accumulator  is  attached  to  the  pipe 
system,  and  while  the  press  is  idle  the  pump  slowly  raises  the 
accumulator  weight,  thereby  storing  potential  energy,  which  is 
given  out  rapidly  by  the  descending  weight  for  a  short  time  while 
the  press  acts,  (c)  A  dynamo-electric  machine  is  run  by  a  steam- 
engine,  and  the  electrical  energy  is  delivered  and  stored  in  storage 
batteries,  upon  which  there  is  a  periodical  demand.  In  this  case, 
as  well  as  in  case  (&),  there  is  a  transformation  of  energy  as  well 
as  a  transfer  in  time. 


5  6  MACHINE  DESIGN. 

44.  Force  Problems. — Suppose  the  slider-crank  mechanism  in 
Fig.  40  to  represent  a  shaping-machine,  the  velocity  diagram  of 


FIG.  40. 


the  slider  being  drawn.  The  resistance  offered  to  cutting  metal 
during  the  forward  stroke  must  be  overcome.  This  resistance 
may  be  assumed  consta'nt.  Throughout  the  cutting  stroke  there 
is  a  continually  varying  rate  of  doing  work.  This  is  because  the 
rate- of  doing  work  =  resisting  force  (constant)  X  velocity  (vary- 
ing). This  product  is  continually  varying,  and  is  a  maximum 
when  the  slider's  velocity  is  a  maximum.  The  slider  must  be 
driven  by  means  of  energy  transmitted  through  the  crank  a.  The 
maximum  rate  at  which  energy  must  be  supplied  equals  the  maxi- 
mum rate  of  doing  work  at  the  slider.  Draw  the  mechanism  in 
the  position  of  maximum  velocity  of  slider;  *  i.e.,  locate  the  center 
of  the  slider-pin  at  the  base  of  the  maximum  ordinate  of  the  veloc- 
ity diagram,  and  draw  b  and  a  in  their  corresponding  positions. 
The  slider's  known  velocity  is  represented  by  y,  and  the  crank- 
pin's  required  velocity  is  represented  by  a  on  the  same  scale. 
Hence  the  value  of  a  becomes  known  by  simple  proportion.  The 
rate  of  doing  work  must  be  the  same  at  c  and  at  ab  (neglecting 
friction). |  Hence  Rv\=  Fv2,  in  which  R  and  v\  represent  the 

*  It  is  customary  to  assume  the  slider's  position  for  this  condition  to  be  that 
corresponding  to  an  angle  of  90°  between  crank  and  connecting-rod.  This  is  not: 
exactly  true,  but  is  a  sufficiently  close  approximation  for  the  ordinary  proportions 
of  crank  and  connecting-rod  lengths.  For  method  of  exact  determination  of 
slider's  position  see  Appendix. 

t  The  effect  of  acceleration  to  redistribute  energy  is  zero  in  this  position,  be- 
cause the  acceleration  of  the  slider  at  maximum  velocity  is  zero,  and  the  angular 
acceleration  of  b  can  only  produce  pressure  in  the  journal  at  ad.  If  Ra  equals- 


ENERGY  IN  MACHINES.  57 

force  and  velocity  factors  at  c\  and  F  and  v2  represent  the  tangen- 
tial forre  and  velocity  factors  ab.  R  and  vt  are  known  from  the 
conditions  of  the  problem,  and  v2  is  found  as  above.  Hence  F  may 

be  found,  =——= force  which,  applied  tangentially  to  the  crank- 
pin  center,  will  overcome  the  maximum  resistance  of  the  machine. 
In  all  other  positions  of  the  cutting  stroke  the  rate  of  doing  work 
is  less,  and  F  would  be  less.  But  it  is  necessary  to  provide  driv- 
ing mechanism  capable  of  overcoming  the  maximum  resistance, 
when  no  fly-wheel  is  used.  If  now  F  be  multiplied  by  the  crank 
radius,  the  product  equals  the  maximum  torsional  moment  (  =  M) 
required  to  drive  the  machine.  If  the  energy  is  received  on  some 
different  radius,  as  in  case  of  gear  or  belt  transmission,  the  maxi- 
mum driving  force  =  M  +  the  new  radius.  During  the  return 
stroke  the  cutting-tool  is  idle,  and  it  is  only  necessary  to  overcome 
the  frictional  resistance  to  motion  of  the  bearing  surfaces.  Hence 
the  return  stroke  is  not  considered  in  designing  the  driving  mech- 
anism. When  the  method  of  driving  this  machine  is  decided  on, 
the  capacity  of  the  driving  mechanism  must  be  such  that  it  shall 
be  capable  of  supplying  to  the  crank-shaft  the  torsional  driving 
moment  M,  determined  as  above. 

This  method  applies  as  well  to  the  quick-return  mechanisms 
given.  In  each,  when  the  velocity  diagram  is  drawn,  the  vector 
of  the  maximum  linear  velocity  of  the  slider,  *=Z,i,  and  of  the 
constant  linear  velocity  of  the  crank-pin  center,  =  L2,  are  known, 
and  the  velocities  corresponding,  v\  and  v2,  are  also  known,  from 
the  scale  of  velocities.  The  rate  of  doing  work  at  the  slider  and 

the  force  necessary  to  produce  accelcrati  jn  of  trie  slider  mass  at  any  position  and 
Fa  the  force  necessary  at  the  crank  pin  to  produce  targ~nt:al  acceleration  of  the 
rotating  mass  (assuming  a  variable  velocity  of  the  crank-pin  as  well  as  slider), 
then  the  equation  in  its  most  general  form  will  be  (R+ RO)VI=  (F+ Fa) v2. 
With  uniform  rate  of  rotation  of  the  crank  this  becomes  (R  +  Ra)vl= Fvt) ;  and 
for  position  corresponding  to  maximum  velocity  of  slider,  as  above,  Rvi  =  Fvy 


58  MACHINE  DESIGN. 

at  the  crank-pin  center  is  the  same,  friction  being  neglected. 
Hence  Rvi=Fv2,  or,  since  the  vector  lengths  are  proportional 

-nj 

to  the  velocities  they  represent,  RL\  =  FL^\  andF=-^ — -.     There- 

-Z-/2 

fore  the  resistance  to  the  slider's  motion,  =R,  on  the  cutting 
stroke,  multiplied  by  the  ratio  of  linear  velocity  vectors,  -=4,  of 

L2 

slider  and  crank-pin,  equals  F,  the  maximum  force  that  must  bo 
applied  tangentially  at  the  crank-pin  center  to  insure  motion. 
F  multiplied  by  the  crank  radius  =  maximum  torsional  driving 
moment  required  by  the  crank-shaft.  If  R  is  varying  and  known, 
find  where  Rv,  the  rate  of  doing  work,  is  a  maximum,  and  solve 
for  that  position  in  the  same  way  as  above. 

Where  the  mass  to  be  accelerated  is  considerable  the  maxi- 
mum effort  will  be  called  for  at  the  beginning  of  each  stroke. 
If  there  is  a  quick  return  the  maximum  effort  will  come  at  the 
beginning  of  the  return  stroke.  A  planer  calls  for  about  twice 
as  much  power  at  the  beginning  of  its  return  stroke  as  it  does 
during  its  cutting  stroke. 

45.  Force  Problems,  Continued. — In  the  usual  type  of  steam- 
engine  the  slider-crank  mechanism  is  used,  but  energy  is  supplied 
to  the  slider  (which  represents  piston,  piston-rod,  and  cross- 
head),  and  the  resistance  opposes  the  rotation  of  the  crank  and 
attached  shaft.  In  any  position  of  the  mechanism  (Fig.  41), 
force  applied  to  the  crank-pin  through  the  connecting-rod  may 
be  resolved  into  two  components,  one  radial  and  one  tangential. 
The  tangential  component  tends  to  produce  rotation;  the  radial 
component  produces  pressure  between  the  surfaces  of  the  shaft - 
journal  and  its  bearing.  The  tangential  component  is  approxi- 
mately a  maximum  when  the  angle  between  crank  and  connecting- 
rod  equals  90°,*  and  it  becomes  zero  when  C  reaches  A  or  B. 
If  there  is  a  uniform  resistance  the  rate  of  doing  work  is  constant. 

*  See  foot-note  on  page  56. 


ENERGY  IN  MACHINES.  59 

Hence,  since  the  energy  is  supplied  at  a  varying  rate,  it  follows 
that  during  part  of  the  revolution  the  effort  is  greater  than  the 
resistance;  while  during  the  remaining  portion  of  the  revolution 
the  effort  is  less  than  the  resistance,  and  the  machine  will  stop 
unless  other  means  are  provided  to  maintain  motion.  A  "fly- 
wheel "  is  keyed  to  the  shaft,  and  this  wheel,  because  of  slight 


FIG.  41. 

variations  of  velocity,  alternately  stores  and  gives  out  the  excess 
and  deficiency  of  energy  of  the  effort,  thereby  adapting  it  to  the 
constant  work  to  be  done.* 

46.  Problem.  —  Given  length  of  stroke  of  the  slider  of  a  steam- 
engine  slider-crank  mechanism,  the  required  horse-power,  01 
rate  of  doing  work,  and  number  of  revolutions.  Required  the 
total  mean  pressure  that  must  be  applied  to  the  piston. 

Let       L  =  length  of  stroke  =  i  foot; 
HP  =  horse-power  =  20; 
N  =  strokes  per  minute  =200; 
F  =  required  mean  force  on  piston. 
Then  N  XL  =  200  feet  per  minute  =  mean  velocity  of  slider  =  F. 

Now,  the  mean  rate  of  doing  work  in  the  cylinder  and  at  the 
main  shaft  during  each  stroke  is  the  same  (friction  neglected)  > 
hence  FV=HPx  33000, 


See  Chapter  XVI. 


6o 


MACHINE  DESIGN. 


47.  Solution  of  Force  Problem  in  the  Slider-crank  Chain. 
— In  the  slider-crank  chain  the  velocity  of  the  slider  necessarily 
varies  from  zero  at  the  ends  of  its  stroke  to  a  maximum  value 
near  mid-stroke.  The  mass  of  the  slider  and  attached  parts 
is  therefore  positively  and  negatively  accelerated  each  stroke. 
When  a  mass  is  positively  accelerated  it  stores  energy;  and 
when  it  is  negatively  accelerated  it  gives  out  energy.  The  amount 
of  this  energy,  stored  or  given  out,  depends  upon  the  mass  and 
the  acceleration.  The  slider  stores  energy  during  the  first  part 
of  its  stroke  and  gives  it  out  during  the  second  part  of  its  stroke. 


O  m 

<LC 


\l 


FIG.  42. 

While,  therefore,  it  gives  out  all  the  energy  it  receives,  it  gives 
it  out  differently  distributed.  In  order  to  find  exactly  how  the 
energy  is  distributed,  it  is  necessary  to  find  the  acceleration 
throughout  the  slider's  stroke.  This  may  be  done  as  follows: 
Fig.  42,  A,  shows  the  velocity  diagram  of  the  slider  of  a  slider- 
crank  mechanism  for  the  forward  stroke,  the  ordinates  repre- 
senting velocities,  the  corresponding  abscissae  representing  the 

slider  positions.  The  acceleration  required  at  any  point  =^-, 
in  which  Av  is  the  increase  in  velocity  during  any  interval  of 


ENERGY  IN  MACHINES.  61 

time  At,  assuming  that  the  increase  in  velocity  becomes  constant 
at  that  point.  Lay  off  the  horizontal  line  OP=MN.  Divide 
OP  into  as  many  equal  parts  as  there  are  unequal  parts  in  MN. 
These  divisions  may  each  represent  At.  At  m  erect  the  ordinate 
mn=m\n\,  and  at  o  erect  the  ordinate  op  =  o\pi.  Continue 
this  construction  throughout  OP,  and  draw  a  curve  through  the 
upper  extremities  of  the  ordinates.  Fig.  42,  B,  is  a  velocity 
diagram  on  a  "time  base."  At  O  draw  the  tangent  OT  to  the 
curve.  If  the  increase  in  velocity  were  uniform  during  the  time 
interval  represented  by  Om,  the  increment  of  velocity  would  be 
represented  by  mT.  Therefore  mT  is  proportional  to  the  accel- 
eration at  the  point  O,  and  may  be  laid  off  as  an  ordinate  of  an 
acceleration  diagram  (Fig.  426").  Thus  Qa=mT.  The  divi- 
sions of  QR  are  the  same  as  those  of  MN;  i.e.,  they  represent. 
positions  of  the  slider.  This  construction  may  be  repeated  for 
the  other  divisions  of  the  curve  B.  Thus  at  n  the  tangent  nTi 
and  horizontal  nq  are  drawn,  and  qT\  is  proportional  to  the 
acceleration  at  n,  and  is  laid  off  as  an  ordinate  be  of  the  ac- 
celeration diagram.  To  find  the  value  in  acceleration  units 
of  Qa,  mT  is  read  off  in  velocity  units  =  Av  by  the  scale 
of  ordinates  of  the  velocity  diagram.  This  value  is  divided 
by  At,  the  time  increment  corresponding  to  Om.  The  result 

Av 
of  this  division   -j-  =  acceleration   at   M  in   acceleration   units. 

J/  =  the  time  of  one  stroke,  or  of  one  half  revolution  of  the  crank 
divided  by  the  number  of  divisions  in  OP.  If  the  linear  velocity 
of  the  center  of  the  crank-pin  in  feet  per  second,  =v,  be  repre- 
sented by  the  length  of  the  crank  radius  = =  a,  then  the  scale 

of  velocities,  or  velocity  in  feet  per  second  for  i  inch  of  ordinate, 

=  — =  — 7 — .     D  is  the  actual  diameter  of  the  crank  circle,  N 
a      aoo 

is  the  number  of  revolutions  per  minute,  and  a  is  the  crank  radius 
measured  on  the  figure. 


62  MACHINE  DESIGN. 

The  determination  of  the  acceleration  curve,  by  means  of 
tangents  drawn  to  the  "time- base"  velocity  curve,  has  a  serious 
drawback.  The  tangent  lines  are  laid  down  by  inspection,  and  slight 
inaccuracy  in  their  location  and  construction  may  lead  to  consid- 
erable errors  in  the  ordinates  obtained  for  the  acceleration  curve. 

The  following  method  is  therefore  suggested  as  an  alternative. 

If  one  point  is  rotating  about  another  point  with  a  given 
instantaneous  velocity =v  and  a  radius  =  r,  the  instantaneous 

v2 
radial  acceleration  of  either  point  toward  the  other=  — . 

Consider  the  slider-crank  chain  in  the  position  at  the  begin- 
ning of  the  forward  stroke  as  shown  in  Fig.  43^.  The  problem 
is  to  determine  the  acceleration  of  the  point  be  toward  ad.  Ac- 
celerations toward  the  right  will  be  considered  as  positive,  toward 
the  left  as  negative.  In  the  position  chosen  the  point  ab  is  mov- 
ing, relatively  to  both  links  d  and  c,  in  the  direction  of  the  arrow 
with  a  velocity =v,  the  uniform  velocity  of  ab  relatively  to  d. 

The  acceleration  of  be  toward  ad  is  always  made  up  of  two 
components,  namely,  the  acceleration  of  be  toward  ab  and  the 
acceleration  of  ab  toward  ad.     In  the  position  under  considera- 
te2 
tion  the  acceleration  of  be  toward  ab=-r  in  a  positive  direction. 

v2 

Similarly  the  acceleration  of  ab  toward  ad=—  in  a  positive  di- 
rection.    The  total  acceleration  of  be  toward  ad  therefore  equals 

v2    v2 
the  sum  of  these  two  components,  or  =  -7-  +  —. 

On  the  other  hand,  at  the  end  of  the  forward  stroke,  shown 

v2 
in  Fig.  43^,  the  acceleration  of  be  toward  d>  =  -r  in  a  positive 

j.2 

direction  as  before,  while  the  acceleration  of  ab  toward  ad= — 

a 

in  a  negative  direction.     The  algebraic  sum  of  these  two  com- 


ENERGY  IN  MACHINES. 


ponents  therefore  =  -r  —  — .     This   quantity  will  always  have  a 

negative  value,  since  in  the  slider-crank  mechanism  a  must  al- 
ways be  smaller  than   b. 

To  construct  the  acceleration  curve,  lay  off  a  length  MN 


FIG.  44. 

(Fig.  43Q  proportionate  to  the  length  of  the  stroke  of  the  slider. 
At  M  erect  an  ordinate,  MP,  whose  value  equals  -r  +  — .  It 
is  best  to  use  for  these  ordinates  a  scale  on  which  a  (the  length 


04  MACHINE  DESIGN. 

V2 

of  the  crank)  represents  the  value  — .     At  N  erect  the  negative 

v2    v2 

ordinate  NQ=-r— — .* 
b      a 

There  is  a  position  of  the  slider,  O,  where  the  acceleration 
equals  zero.     This  must  correspond  to  the  position  of  the  slider 

*  The  following  construction  for  graphically  obtaining  ordinates  representing 

•  -  +  —  and  7-       — ,  on  the  scale  upon  which  a  represents  v 
b         a          b        a 

(V2\ 
and,    hence,  a  =  —  I  is  due  to  Professor  Le  Conte. 
a) 


FIG.  447. 

Reference  is  to  Fig.  441.  M  and  2V  represent  the  position  of  the  slider  at  the 
beginning  and  end  of  the  stroke,  respectively. 

At  abl  erect  the  vector  v  ( =  a)  and  from  M  draw  a  line  through  its  upper  extremity. 
Prolong  this  line  until  it  cuts  the  perpendicular  through  ad,  thus  determining  the 
length  yr 

At  ab2  lay  off  downward  the  vector  v  (  =  a).  From  2V  draw  a  line  to  the  lower 
extremity  of  this  vector,  cutting  off  the  length  y2  on  the  perpendicular  through  ad 

Then  will  yl  represent  —  H — ;    and  yz  represent  —  — 


For,  taking  slider  position  M,  by  similar  triangles, 


a2  +  ab 


=  u-  +  a.       But  a  =  "-,    .-.  ft  -  r  +  - 
b  aba 


For  position  2V,  by  similar  triangles,  —^-:  =  —^7 — ' 

a2  -    ab      v2  _  r==7/_2  _^2 
•*•  y*~        b        ~b   ~a~   b      a  ' 


ENERGY  IN  MACHINES.  65 

when  it  has  its  maximum  velocity,  which  may  be  taken  from 
the  original  velocity  diagram  of  the  slider,  or,  with  greater 
accuracy,  from  Curve  B  in  the  Appendix.  Through  POQ  draw 
a  smooth  curve.  For  most  purposes  this  curve  will  be  accurate 
enough. 

Where  more  points  of  the  curve  are  desired  for  the  sake  of 
greater  accuracy  the  method  illustrated  in  Fig.  44  may  be 
used.  Assume  the  slider  in  the  position  at  which  its  acceleration 
is  desired  and  draw  the  crank  a  and  connecting-rod  b  in  their 
corresponding  positions.  Locate  the  centres  ad,  ah,  ac,  and  bd. 
From  ac  draw  a  parallel  to  bc-ad  until  it  cuts  the  crank,  pro- 
longed if  necessary,  at  A.  From  A  draw  a  parallel  to  ad-ac 
until  it  cuts  the  connecting-rod  at  B.  From  B  draw  a  perpen- 
dicular to  the  connecting-rod  until  it  cuts  bc-ad,  prolonged  if 
necessary,  at  C.  Then  ad-C  is  the  desired  ordinate  of  the 

v2 
acceleration  diagram  on  the  scale  by  which  the  length  a  =  — . 

The  proof  is  as  follows,  reference  being  made  to  Fig.  44. 

At  this  instant  every  point  of  b  relatively  to  c  is  swinging 
about  the  centre  be  with  a  velocity  proportional  to  its  distance 
from  be. 

vel.  of  bd  rel.  to  c    bd-bc     ac-ad 
vel.  of  ab  rel.  to  c     ab-bc    ac-ab' 

But  ac-ad  represents  the  velocity  of  c  relatively  to  d  (or  d 
relatively  to  c)  on  the  same  scale  that  ad-ab  represents  the 
velocity  of  the  point  ab  relatively  to  d.  Therefore  ac-ab  repre- 
sents the  velocity  of  ab  rotating  about  be  relatively  to  c  on  the  same 
scale  that  ad-ab  represents  the  velocity  of  ab  relatively  to  d. 

Hence  the  radial  acceleration  of  ab  toward  be  *  (or  conversely 

*  The  acceleration  of  a  point  A  with  respect  to  another  point  B  is  the  accel- 
eration of  A  with  respect  to  a  non-rotating  body  of  which  5  is  a  point. 


66  MACHINE  DESIGN. 


, 

of  be  toward  ab)  =  —  7  —  ,  which  is   represented  by  the  length 

ab-B,  as  can  be  shown  as  follows: 
By  similar  triangles 

ab-B     ab-A      ab-ac 
ab-ac  ~  ab-ad  ~  ab-bc 


.    7,     i?     ab-ac      ab-ac 
.  ba-B  = 


ab-bc          b 

ab—aa 

The   radial   acceleration    of   ab    toward     ad=  --  ,  whose 

a 

value  we  represent  by  the  length  a.  The  component  of  this 
acceleration  in  the  direction  bc-ab=ab-D. 

The  acceleration  of  be,  relatively  to  d,  along  the  path  bc-ab 
is  made  up  of  two  components:  ist,  the  acceleration  of  be  toward 
ab(=B-ab)  plus,  2d,  the  acceleration  of  ab  relatively  to  d  along 
the  same  path  (=ab-D). 

In  the  position  shown  this  algebraic  sum  is  the  negative 
quantity  represented  by  B-D.  But  the  actual  direction  of  fee's 
acceleration  relatively  to  d  is  along  the  line  bc-ad.  Its  accelera- 
tion in  this  direction  must  therefore  be  the  quantity  whose  com- 
ponent along  ab-bc  is  B-D,  namely,  C-ad.  Q.E.D. 

If  the  weight  W  of  parts  accelerated  is  known,  the  force  F 
necessary  to  produce  the  acceleration  at  any  slider  position  may 
be  found  from  the  fundamental  formula  of  mechanics, 


p  being  the  acceleration  corresponding  to  the  position  considered. 
If  the  ordinates  of  the  acceleration  diagram  are  taken  as  repre- 
senting the  forces  which  produce  the  acceleration,  the  diagram 
will  have  force  ordinates  and  space  abscissae,  and  areas  will 
represent  work.  Thus,  Qas,  Fig.  426",  represents  the  work  stored 


ENERGY  IN  MACHINES.  67 

during  acceleration,  and  Rsd  represents  the  work  given  out  during 
retardation.  Let  MN,  Fig.  45,  represent  the  length  of  the 
slider's  stroke  and  NC  the  resistance  of  cutting  (uniform)  on  the 
same  force  scale  as  that  by  which  Qa,  Fig.  426*,  represents  the 

Wp 
force  — —  at  the  beginning  of  the  stroke;  then  energy  to  do  cutting 

O 

per  stroke  is  represented  by  the  area  MBCN.  But  during  the 
early  part  of  the  stroke  the  reciprocating  parts  must  be  acceler- 
ated, and  the  force  necessary  at  the  beginning,  found  as  above, 
=  BD=Qa.  The  driving-gear  must,  therefore,  be  able  to  over- 
come resistance  equal  to  MB  +  BD.  The  acceleration,  and  hence 
the  accelerating  force,  decreases  as  the  slider  advances,  becoming 
zero  at  E.  From  E  on  the  acceleration  becomes  negative,  and 
hence  the  slider  gives  out  energy  and  helps  to  overcome  the  resist- 
ance, and  the  driving-gear  has  only  to  furnish  energy  represented 
by  the  area  AEFN,  though  the  work  really 
|C  done  against  resistance  equals  that  repre- 
A  sented  by  the  area  CEFN.  The  energy 
represented  by  the  difference  of  these  areas, 
=ACE,  is  that  which  is  stored  in  the 
slider's  mass  during  acceleration.  Since  by  the  law  of  con- 
servation of  energy,  energy  given  out  per  cycle=that  received, 
it  follows  that  area  4CE=area  DEB,  and  area  BCMN  = 
ADMN.  This  redistribution  of  energy  would  seem  to  modify 
the  problem  on  page  52,  since  that  problem  is  based  on  the 
assumption  of  uniform  resistance  during  cutting  stroke.  The 
position  of  maximum  velocity  of  slider,  however,  corresponds  to 
acceleration  =o.  The  maximum  rate  of  doing  work,  and  the 
corresponding  torsional  driving  moment  at  the  crank-shaft  would 
probably  correspond  to  the  same  position,  and  would  not  be 
materially  changed.  In  such  machines  as  shapers,  the  accelera- 
tion and  weight  of  slider  are  so  small  that  the  redistribution  of 
energy  is  unimportant. 


68 


MACHINE  DESIGN. 


48.  Solution  of  the  Force  Problem  in  the  Steam-engine  Slider- 
crank  Mechanism.  (Slider  represents  piston  with  its  rod,  and  the 
cross-head.)  —  The  steam  acts  upon  the  piston  with  a  pressure 
which  varies  during  the  stroke.  The  pressure  is  redistributed 
before  reaching  the  cross-head  pin,  because  the  reciprocating  parts 
are  accelerated  in  the  first  part  of  the  stroke,  with  accompanying 
storing  of  energy  and  reduction  of  pressure  on  the  cross-head 
pin;  and  retarded  in  the  second  part  of  the  stroke,  with  accom- 
panying giving  out  of  energy  and  increase  of  pressure  on  the 
cross-head  pin.  Let  the  ordinates  of  the  full  line  diagram  above 
OX,  Fig.  46,4,  represent  the  total  effective  pressure  on  the  piston 
throughout  a  stroke.  Fig.  46-8  is  the  velocity  diagram  of  slider. 


FIG.  46. 

Find  the  acceleration  throughout  stroke,  and  from  this  and  the 
known  value  of  weight  of  slider  find  the  force  due  to  acceleration. 
Draw  diagram  Fig.  466",  whose  ordinates  represent  the  force 
due  to  acceleration,  upon  the  same  force  scale  used  in  A.  Lay 
off  this  diagram  on  OX  as  a  base  line,  thereby  locating  the  dotted 
line.  The  vertical  ordinates  between  this  dotted  line  and  the 
upper  line  of  A  represent  the  pressure  applied  to  the  cross-head 
pin.  These  ordinates  may  be  laid  off  from  a  horizontal  base  line, 
giving  D.  The  product  of  the  values  of  the  corresponding  ordinates 
of  B  and  D  =the  rate  oj  doing  work  throughout  the  stroke.  Thus 
the  value  of  GH  in  pounds  X  value  of  EF  in  feet  per  second  =  the 


ENERGY  IN  MACHINES.  69 

rate  of  doing  work  in  foot-pounds  per  second  upon  the  cross- 
head  pin,  when  the  center  of  the  cross-head  pin  is  at  E.  The 
rate  of  doing  work  at  the  crank-pin  is  the  same  as  at  the  cross- 
head  pin.  Hence  dividing  this  rate  of  doing  work,  =EFxGH, 
by  the  constant  tangential  velocity  of  the  crank-pin  center,  gives 
the  force  acting  tangentially  on  the  crank-pin  to  produce  rotation. 
The  tangential  forces  acting  throughout  a  half  revolution  of 
the  crank  may  be  thus  found,  and  plotted  upon  a  horizontal 
base  line=length  of  half  the  crank  circle  (Fig.  47.8).  The  work 
done  upon  the  piston,  cross-head  pin,  and  crank  during  a  piston 
stroke  is  the  same.  Hence  the  areas  of  A  and  D,  Fig.  46,  are 
equal  to  each  other,  and  to  the  area  of  the  diagram,  Fig.  4jB. 
The  forces  acting  along  the  connecting-rod  for  all  positions 
during  the  piston  stroke  may  be  found  by  drawing  force  triangles 
with  one  side  horizontal,  one  vertical,  and  one  parallel  to  position 
of  connecting-rod  axis,  the  horizontal  side  being  equal  to  the 
corresponding  ordinate  of  Fig.  46D.  The  vertical  sides  of  these 
triangles  will  represent  the  guide  reaction,  while  the  side  parallel 
to  the  connecting-rod  axis  represents  the  force  transmitted  by 
the  connecting-rod. 

The  tangential  forces  acting  on  the  crank-pin  may  be  found 
graphically  by  the  method  shown  in  Fig.  47^4.  Let  GH  repre- 
sent the  net  effective  force  acting  in  a  horizontal  direction  at  the 
center  of  the  cross-head  pin. 

It  has  been  shown  that  EF  represents  the  velocity  of  the 
slider  on  the  same  scale  that  EA  represents  that  of  the  center  of 
the  crank-pin;  also  that  the  rate  of  doing  work,  after  having 
made  the  necessary  corrections  for  acceleration,  is  the  same  at 
the  center  of  the  crank-pin  as  at  the  slider,  i.e.,  GHxEF  =  tan- 
gential  force  at  center  of  crank-pin  XEA.  Hence  the  tangential 

J7  77 

force  at  center  of  crank-pin  =  GH  X  pi  • 


7o 


MACHINE  DESIGN. 


Lay  off  AB=GH,  and  draw  BC  parallel  to  EF.     Then,  by 
similar  triangles, 

BC     EF     .  EF  EF 


the  tangential  force  acting  at  the  crank-pin  center  for  the  assumed 
position  of  the  mechanism,  on  the  same  scale  as  G.ff=net  effective 
horizontal  force  on  slider. 

Lay  off  AD=BC.      . 

Following  through  this  construction  for  a  number  of  positions 
of  the  mechanisms,  a  polar  diagram  is  determined  which  shows 


FIG.  478. 

very  clearly  the  relation  existing  between  the  varying  tangential 
forces  and  the  corresponding  crank  positions.  Before  this  dia- 
gram may  be  used  in  the  solution  of  the  fly-wheel  problem  (see 
Chapter  XVI)  it  should  be  transferred  to  a  straight-line  base 
whose  length  for  one  stroke  equals  the  semi-circumference  of  the 
crank-pin  circle.  That  is,  the  abscissae  will  be  the  distance  moved 
through  by  the  center  of  the  crank-pin  and  the  ordinates  will  be 
the  corresponding  radial  intercepts  AD.  The  diagram  so  ob- 
tained will  be  identical  with  that  shown  in  Fig.  47.5. 


CHAPTER  VI. 

PROPORTIONS   OF  MACHINE   PARTS   AS   DICTATED   BY  STRESS. 

49.  The  size  and  form  of  machine '  parts  *  are  governed  by 
six  main  considerations : 

(1)  The  size  and  nature  of  the  work  to  be  accommodated  (as 
the  swing  of  engine-lathes,  etc.). 

(2)  The  stresses  which  they  have  to  endure. 

(3)  The  maintaining  of  truth  and  accuracy  against  wear,  in- 
cluding all  questions  of  lubrication. 

(4)  The  cost  of  production. 

(5)  Appearance. 

(6)  Properties  of  materials  to  be  used. 

The  first  is  a  given  condition  in  any  problem ;  the  second  will 
be  discussed  here;  the  third  will  be  treated  in  the  chapters  on 
Journals  and  Sliding  Surfaces;  the  fourth  is  touched  upon  here; 
the  principles  governing  the  fifth  are  treated  in  Chapter  XIX 
and  here. 

It  is  assumed  in  this  and  following  chapters  that  the  reader  is 
familiar  with  the  properties  of  the  materials  employed  in  machine 
construction, f  and  with  the  general  principles  of  the  science  of 
mechanics. 

50.  The  stresses  acting  on  machine  parts  may  be  constant, 
variable,  or  suddenly  applied. 


*  On  this  general  subject  see  an  excellent  article  by  Prof.  Sweet  in  the  Jour- 
nal of  the  Franklin  Institute,  3d  Series,  Vol.  125,  pp.  278-300.  The  reader  is 
also  referred  to  "A  Manual  of  Machine  Construction,"  by  Mr.  John  Richards, 
and  to  the  Introduction  of  this  volume. 

f  See  Smith's  "Materials  of  Machines." 

71 


72  MACHINE  DESIGN. 

A  CONSTANT  stress  is  frequently  spoken  of   as  a  STEADY,  or 

DEAD,  LOAD. 

A  VARIABLE  stress  is  known  as  a  LIVE  LOAD. 

A  SUDDENLY  APPLIED  stress  is  known  as  a  SHOCK. 

51.  Constant  Stress. — If  a  machine  part  is  subjected  to  a  con- 
stant stress,  i.e.,  an  unvarying  load  constantly  applied,  its  design 
becomes  a  simple  matter,  as  the  amount  of  such  a  stress  can 
generally  be  very  closely  estimated.  Knowing  this  and  the 
properties  of  the  materials  to  be  used,  it  is  only  necessary  to  cal- 
culate the  area  which  will  sustain  the  load  without  excessive 
deformation. 

Thus,  in  simple  tension  or  compression,  if  we  let  £7= the  ulti- 
mate strength  of  the  material  in  pounds  per  square  inch,  F  the 
total  constant  stress  in  pounds,  A  the  unknown  area  in  square 
inches  necessary  to  sustain  F,  we  write 


where  K  is  a  so-called  FACTOR  OF  SAFETY,  introduced  to  reduce 
the  permitted  unit  stress  to  such  a  point  as  will  limit  the  deforma- 
tion (strain)  to  an  allowable  amount,  and  also  to  provide  for  pos- 
sible defects  in  the  material  itself.  In  exceptional  cases  where 
the  stresses  permit  of  accurate  calculation,  and  the  material  is  of 
proven  high  grade  and  positively  known  strength,  K  has  been 
given  as  low  a  value  as  i-£;  but  values  of  2  and  3  are  ordinarily 
used  for  wrought  iron  and  steel  free  from  welds;  while  4  to  5  are 
as  small  as  should  be  used  for  cast  iron,  on  account  of  the  uncer- 
tainty of  its  composition,  the  danger  of  sponginess  of  structure, 
and  indeterminate  shrinkage  stresses. 

The  SAFE  UNIT  STRESS  =/  =  T?  in  pounds  per  square  inch. 
A. 

52.  Variable  Stress.  — We  pass  next  to  the  consideration  of 


PROPORTIONS  OF  MACHINE  PARTS  AS  DICTATED  BY  STRESS.  73 

variable  stresses  or  live  loads.  Here  the  problem  is  much  more 
complex  than  with  dead  loads. 

Experiments  by  Wohler,*  and  Bauschinger,f  with  the  work 
of  Weyrauch  |  and  others  have  given  us  the  laws  of  bodies  sub- 
jected to  repeated  stresses.  In  substance  Wohler's  law  is  as 
follows:  MATERIAL  MAY  BE  BROKEN  BY  REPEATED  APPLICATIONS 

OF  A  FORCE  WHICH  WOULD  BE  INSUFFICIENT  TO  PRODUCE  RUPTURE 
BY  A  SINGLE  APPLICATION.  THE  BREAKING  IS  A  FUNCTION  OF 
RANGE  OF  STRESS ;  AND  AS  THE  VALUE  OF  THE  RECURRING  STRESS 
INCREASES,  THE  RANGE  NECESSARY  TO  PRODUCE  RUPTURE  DE- 
CREASES. IF  THE  STRESS  BE  REVERSED,  THE  RANGE  EQUALS 
THE  SUM  OF  THE  POSITIVE  AND  NEGATIVE  STRESS. 

Bauschinger's  conclusions  were  as  follows: 

(1)  WlTH   REPEATED   TENSILE   STRESSES   WHOSE   LOWER    LIMIT 
WAS    ZERO,    AND   WHOSE    UPPER   LIMIT   WAS   NEAR  THE    ORIGINAL 
ELASTIC    LIMIT,    RUPTURE    DID    NOT    OCCUR   WITH    FROM    5    TO    l6 

MILLION  REPETITIONS.  He  cautions  the  designer  (a)  that  this 
will  not  hold  for  DEFECTIVE  material,  i.e.,  a  factor  of  safety  must 
still  be  used  for  this  reason;  and  (b)  that  the  elastic  limit  of  the 
material  must  be  carefully  determined,  because  it  may  have  been 
artificially  raised  by  cold  working,  in  which  case  it  does  not  accur- 
ately represent  the  material.  The  original  elastic  limit  may  be  de- 
termined by  testing  a  piece  of  the  material  after  careful  annealing. 

(2)  WlTH  OFTEN- REPEATED  STRESSES  VARYING  BETWEEN  ZERO 
AND  AN    UPPER  STRESS  WHICH    IS    IN  THE  NEIGHBORHOOD  OF  OR 
ABOVE  THE  ELASTIC  LIMIT,   THE  LATTER  IS  RAISED  EVEN  ABOVE, 
OFTEN  FAR  ABOVE,  THE  UPPER  LIMIT  OF  STRESS,  AND  IT  IS  RAISED 
HIGHER  AS  THE  NUMBER  OF   REPETITIONS   OF   STRESS   INCREASES, 

*  "Ueberdie  Festigkeitsversuche  mit  Eisen  und  Stahl,"  A.  Wohler,  Berlin,  1870. 

f "  Mittheilungen  der  Konig'.  Tech.  Hochschule  zu  Miinchen,"  J.  Bau- 
schinger,  Munich,  1886  and  1897.  . 

J  "Structures  of  Iron  and  Steel,"  by  J.  Weyrauch.  Trans,  by  A.  J.  DuBois, 
New  York,  1877 


74  MACHINE  DESIGN. 

WITHOUT,  HOWEVER,  A  KNOWN  LIMITING  VALUE,  L,  BEING  EX- 
CEEDED. 

(3)  REPEATED  STRESSES  BETWEEN  ZERO  AND  AN  UPPER  LIMIT 

BELOW  L  DO  NOT  CAUSE  RUPTURE;  BUT  IF  THE  UPPER  LIMIT  IS 
ABOVE  L  RUPTURE  WILL  OCCUR  AFTER  A  LIMITING  NUMBER  OF 
REPETITIONS. 

From  this  it  will  be  seen  that  keeping  within  the  ORIGINAL 
ELASTIC  LIMIT  insures  safety  against  rupture  from  repeated 
stress  if  the  stress  is  not  reversed;  and  that,  when  the  stress  is 
reversed,  the  total  range  should  not  exceed  the  ORIGINAL  ELAS- 
TIC RANGE  of  the  material. 

Various  formulae  have  been  proposed  by  different  authorities 
embodying  the  foregoing  laws. 

Unwin's  is  here  given  as  being  most  simple  and  general: 

Let  U  be  the  breaking  strength  of  the  material  in  pounds 
per  square  inch  for  a  load  once  gradually  applied. 

Let  /  max.  be  the  breaking  strength  in  pounds  per  square  inch 
for  the  same  material  subjected  to  a  variable  load  ranging  be- 
tween the  limits  /  max.  and  /  min.,  and  repeated  an  indefinitely 
great  number  of  times,  /min.  is  +  if  the  stress  is  of  the  same 
kind  as  /  max.,  and  is  —  if  the  stress  is  of  the  opposite  kind, 
and  it  is  supposed  that  /  min.  is  not  greater  than  /  max.  Then 
the  range  of  stress  is  J=/max.:F/min.,  the  upper  sign  being 
taken  if  the  stresses  are  of  the  same  kind  and  the  lower  if  they 
are  different.  Hence  J  is  always  positive.  The  formula*  is 


/  max.  =  —  +VU2  -  rj  4U, 


where  t)  is  a  variable  coefficient  whose  value  has  been  experi- 
mentally determined.  For  ductile  iron  and  steel  9  =  1.5,  in- 
creasing with  hardness  of  the  material  to  a  value  of  2. 

*  Unwin's  "Machine  Design,"  Vol.  i,  1903,  pp.  32-36. 


PROPORTIONS  OF  MACHINE  PARTS  AS  DICTATED  BY  STRESS.    75 

This  formula  is  of  general  application. 
Three  cases  may  be  considered: 

(1)  A  constant  stress,  or  dead  load.     In  this  case  the  range 
of  stress,  J  =  o,  and  consequently  /  max.  =  £7,  as  it  should  be. 

(2)  The  stress  is  variable  between  an  upper  limit  and  zero, 
but  is  not  reversed. 

Here  J  =  /  max.,  since  /min.=o,  and  consequently  /  max.  = 


(3)  The  stress  is  reversed,  being  alternately  a  compressive 
and  tensile  stress  of  the  same  magnitude. 

Here        •    /  min.  =  —  /  max.     and     A  =  2}  max. 


/.  /max.= — U. 

2T) 


In  each  case  it  is  necessary  to  divide  the  breaking  load,  /  max., 
by  a  factor  of  safety  in  order  to  get  the  safe  unit  stress  /,  i.e., 

I  = '    ^x\    K  is  a  factor  of  safety  whose  numerical  value  depends 

upon  the  material  used.     (See  Sec.  51.) 

53.  Problem. — Consider  that  there  are  three  pieces  to  be 
designed  using  machinery  steel  having  an  ultimate  tensile  strength 
of  60,000  Ibs.  per  square  inch. 

The  first  piece  sustains  a  steady  load  having  a  dead  weight 
suspended  from  it. 

The  second  piece  is  a  member  of  a  structure  which  is  alter- 
nately loaded  and  unloaded  without  shock. 

The  third  piece  is  subjected  to  alternate  stresses  without 
shock. 

In  each  case  the  maximum  load  is  the  same,  being  30,000  Ibs. 
=  F.  This  material  is  generally  reliable  and  uniform  in  quality. 
A  factor  of  safety  of  3  is  common;  .'.  ^"  =  3  in  each  case;  9  =  1.5. 


76  MACHINE  DESIGN. 

CASE  I. 

/max. 
/max.  =  £7    and     /-- . 

3 

U    60000 
. .  /  =  —  = =  20,000  IDS.  per  sq.  in. 

O  O 

The  necessary  area  A  to  sustain  F  is  determined  by  the  equation 


F      30000 
A  =-r=  — =  ii  sq.  in. 

f         20000         f     » 


CASE 


/  max.  =  2  (v      + 1  -  9)  Z7. 
ij  =  1.5     and     U  =  60,000  Ibs.  per  sq.  in. 
.'.  /  max.  =.6054^7  =  36,324  Ibs.  per  sq.  in. 

/  max. 
/= =  12,108  Ibs.  per  sq.  in. 


F    30000 

A  =7  =7^Q = 2£ 5?-  w->  ««^y- 


CASE  III. 


60000 
.*.  /  max.  =  -  —        =  20,000  Ibs.  per  sq.  in. 

/  max. 
/  =  -  =6667  Ibs.  per  sq.  in. 

F      30000 


The  importance  of  considering  the  question  of  range  of  stress 
in  designing  is  brought  home  by  this  illustration.     Comparison 


PROPORTIONS  OF  MACHINE  PARTS  AS  DICTATED  BY  STRESS.    77 

of  results  shows  that  WITH  THE  SAME  MAXIMUM  LOAD  in  each 
case,  the  second  piece  must  be  given  nearly  twice  the  area  of  the 
first,  while  the  third  must  be  three  times  as  great  in  area  as  the 
first,  the  only  difference  in  the  three  cases  being  the  range  of 
stress. 

54.  Shock.  —  Consideration  of  the  design  of  parts  subjected 
to  shocks  or  suddenly  applied  loads. 

(1)  A  load  is  applied  on  an  unstrained  member  in  a  single 
instant,  but  without  velocity. 

In  this  case,  if  the  stress  does  not  exceed  the  limit  of  elasticity 
of  the  material,  the  stress  produced  will  be  just  twice  that  pro- 
duced by  a  gradually  applied  load  of  the  same  magnitude.  If 
F  =  maximum  total  load  as  before,  then  the  maximum  total 
stress  =  2.F.  The  design  of  the  member  is  then  made  as  in  Case 
II  or  Case  III  of  the  preceding  section. 

(2)  A  load  is  applied  on  an  unstrained  member  in  a  single 
instant,  but  with  velocity. 

In  this  case  the  stress  on  the  member  will  exceed  that  due  to 
a  gradually  applied  load  of  the  same  magnitude  by  an  amount 
depending  on  the  energy  possessed  by  the  load  at  the  moment 
of  impingement. 

Assume  that  a  member  is  stressed  by  a  load  F  falling  through 
a  height  h.  The  unknown  area  of  the  member  =A,  and  the 
allowable  strain  (i.e.,  extension  or  compression)  =  A.  As  before, 
/=  allowable  unit  stress  (determined  by  the  question  of  range  by 
the  use  of  Unwin's  formula). 

The  energy  of  the  falling  load  is 


The  work  done  in  straining  the  member  an  amount  A  with  a 
maximum    fiber    stress  /  is  -XA,  provided  the  elastic  limit  is 


78  MA 'CHINE  DESIGN. 

not  exceeded.     Equating  these  values  of  energy  expended  and 

work  done  and  solving  for  A  gives  A  = 7^ . 

I* 

55.  Form  Dictated  by  Stress.    Tension. — Suppose  that  A  and 
B,  Fig.  48,  are  two  surfaces  in  a  machine  to  be  joined  by  a  member 

subjected  to  simple  tension.    What  is  the  proper 

pi HJ  form   for  the  member?     The  stress  in  all  sec- 

***  tions  of  the  member  at  right  angles  to  the  line 
of  application,  A  B,  of  the  force  will  be  the  same. 
Therefore  the  areas  of  all  such  sections  should  be  equal;  hence 
the  outlines  of  the  members  should  be  straight  lines  parallel  to 
AB.  The  distance  of  the  material  from  the  axis  AB  has  no 
effect  on  its  ability  to  resist  tension.  Therefore  there  is  nothing 
in  the  character  of  the  stress  that  indicates  the  form  of  the 
cross-section  of  the  member.  The  form  most  cheaply  produced, 
both  in  the  rolling-mill  and  the  machine-shop,  is  the  cylindrical 
form.  Economy,  therefore,  dictates  the  circular  cross-section. 
After  the  required  area  necessary  for  safely  resisting  the  stress  is 
determined,  it  is  only  necessary  to  find  the  corresponding  diam- 
eter, and  it  will  be  the  diameter  of  all  sections  of  the  required 
member  if  they  are  made  circular.  Sometimes  in  order  to  get  a 
more  harmonious  design,  it  is  necessary  to  make  the  tension 
member  just  considered  of  rectangular  cross-section,  and  this  is 
allowable  although  it  almost  always  costs  more.  The  thin,  wide 
rectangular  section  should  be  avoided,  however,  because  of  the 
difficulty  of  insuring  a  uniform  distribution  of  stress.  A  unit 
stress  might  result  from  this  at  one  edge  greater  than  the  strength 
of  the  material,  and  the  piece  would  yield  by  tearing,  although 
the  AVERAGE  stress  might  not  have  exceeded  a  safe  value. 

56.  Compression. — If  the   stress  be   compression   instead   of 
tension,  the  same  considerations  dictate  its  form  as  long  as  it 
is  a  "short  block,"  i.e.,  as  long  as  the  ratio  of  length  to  lateral 
dimensions  is  such  that  it  is  sure  to  yield  by  crushing  instead  of 


PROPORTIONS  OF  MACHINE  PARTS  AS  DICTATED  BY  STRESS.   79 

by  "buckling."  A  short  block,  therefore,  should  have  its  longi- 
tudinal outlines  parallel  to  its  axis,  and  its  cross-section  may  be 
of  any  form  that  economy  or  appearance  may  dictate.  Care 
should  be  taken,  however,  that  the  least  lateral  dimension  of 
the  member  be  not  made  so  small  that  it  is  thereby  converted 
into  a  "long  column." 

If  the  ratio  of  longitudinal  to  lateral  dimensions  is  such  that 
the  member  becomes  a  "long  column,"  the  conditions  that  dic- 
tate the  form  are  changed,  because  it  would  yield  by  buckling  or 
flexure  instead  of  crushing.  The  strength  and  stiffness  of  a 
long  column  are  proportional  to  the  moment  of  inertia  of  the 
cross-section  about  a  gravity  axis  at  right  angles  to  the  plane 
in  which  the  flexure  occurs.  A  long  column  with  "fixed  "  or 
"rounded"  ends  has  a  tendency  to  yield  by  buckling  which  is 
equal  in  all  directions.  Therefore  the  moment  of  inertia  needs 
to  be  the  same  about  all  gravity  axes,  and  this  of  course  points  to 
a  circular  section.  Also  the  moment  of  inertia  should  be  as  large 
as  possible  for  a  given  weight  of  material,  and  this  points  to  the 
hollow  section.  The  disposition  of  the  metal  in  a  circular  hollow 
section  is  the  most  economical  one  for  long-column  machine 
members  with  fixed  or  rounded  ends.  This  form,  like  that  for 
tension,  may  be  changed  to  the  rectangular  hollow  section  if 
appearance  requires  such  change.  If  the  long-column  machine 
member  be  "pin  connected,"  the  tendency  to  buckle  is  greatest 
in  a  plane  through  the  line  of  direction  of  the  compressive  force 
and  at  right  angles  to  the  axis  of  the  pins.  The  moment  of  iner- 
tia of  the  cross-section  should  therefore  be  greatest  about  a  gravity 
axis  parallel  to  the  axis  of  the  pins.  Example,  a  steam-engine 
connecting-rod. 

57.  Flexure. — When  the  machine  member  is  subjected  to 
transverse  stress  the  best  form  of  cross-section  is  probably  the  I 
section,  a,  Fig.  49,  in  which  a  relatively  large  moment  of  inertia, 
with  economy  of  material,  is  obtained  by  putting  the  excess  of 


8o  MACHINE  DESIGN. 

the  material  where  it  is  most  effective  to  resist  flexure,  i.e.,  at  the 
greatest  distance  from  the  given  gravity  axis.  Sometimes,  how- 
ever, if  the  I  section  has  to  be  produced  by  cutting  away  the 
material  at  e  and  d,  in  the  machine-shop,  instead  of  producing 
the  form  directly  in  the  rolls,  it  is  cheaper  to  use  the  solid  rect- 
angular section  c.  If  the  member  subjected  to  transverse  stress 
is  for  any  reason  made  of  cast  material,  as  is  often  the  case,  the 
form  b  is  frequently  preferable  for  the  following  reasons: 

(1)  The  best  material  is  almost  sure  to  be  in  the  thinnest  part 
of  a  casting,  and  therefore  in  this  case  at  /  and  g,  where  it  is  most 
effective  to  resist  flexure. 

(2)  The  pattern  for  the  form  b  is  more  cheaply  produced  and 
maintained  than  that  for  a.     The  hollow  box  section,    when 
permitted  by  considerations  of  construction  and  expense  is  still 
better. 

(3)  If  the  surface  is  left  without  finishing  from  the  mold,  any 
imperfections  due  to  the  foundry  work  are  more  easily  corrected 
in  b  than  in  a. 

Machine  members  subjected  to  transverse  stress,  which  con- 
tinually change  their  position  relatively  to  the  force  which  pro- 
duces the  flexure,  should  have  the  same  moment  of  inertia  about 
all  gravity  axes,  as,  for  instance,  rotating  shafts  that  are  strained 
transversely  by  the  force  due  to  the  weight  of  a  fly-wheel,  or  that 
due  to  the  tension  of  a  driving-belt.  The  best  form  of  cross- 
section  in  this  case  is  circular.  The  hollow  section  would  give 
the  greatest  economy  of  material,  but  hollow  members  are  ex- 
pensive to  produce  in  wrought  material,  such  as  is  almost  inva- 
riably used  for  shafts.  The  hollow  circular  section  is  meeting 
with  increasing  use,  especially  for  large  shafts,  on  account  of  the 
combined  lightness  and  strength. 

58.  Torsion. — Torsional  strength  and  stiffness  are  propor- 
tional to  the  polar  moment  of  inertia  of  the  cross-section  of  the 
member.  This  is  equal  to  the  sum  of  the  moments  of  inertia 
about  two  gravity  axes  at  right  angles  to  each  other.  The  forms 


PROPORTIONS  OF  MACHINE  PARTS  AS  DICTATED  BY  STRESS.  8i 

in  Fig.  49  are  therefore  not  correct  forms  for  the  resistance  of 
torsion.  The  circular  solid  or  hollow  section,  or  the  rect- 
angular solid  or  hollow  section,  should  be  used. 

The  I  section,  Fig.  50,  is  a  correct  form  for  resisting  the  stress 
P,  applied  as  shown.  Suppose  the  web  C  to  be  divided  on  the 
line  CD,  and  the  parts  to  be  moved  out  so  that  they  occupy 
the  positions  shown  at  a  and  b.  The  form  thus  obtained  is 
called  a  "box  section."  By  making  this  change  the  moment 
of  inertia  about  ab  has  not  been  changed,  and  therefore  the 
new  form  is  just  as  effective  to  resist  flexure  due  to  the  force  P 
as  it  was  before  the  change.  The  box  section  is  better  able  to 


FIG.  50. 

resist  torsional  stress,  because  the  change  made  to  convert  tne 
I  section  into  the  box  section  has  increased  the  polar  moment 
of  inertia.  The  two  forms  are  equally  good  to  resist  tensile  and 
compressive  force  if  they  are  sections  of  short  blocks.  But  if 
they  are  both  sections  of  long  columns,  the  box  section  would  be 
preferable,  because  the  moments  of  inertia  would  be  more  nearly 
the  same  about  all  gravity  axes. 

59.  Machine  Frames. — The  framing  of  machines  almost  always 
sustains  combined  stresses,  and  if  the  combination  of  stresses 
include  torsion,  flexure  in  different  planes,  or  long-column  com- 
pression, the  box  section  is  the  best  form.  In  fact,  the  box  sec- 
tion is  by  far  the  best  form  for  the  resisting  of  stress  in  machine 
frames.  There  are  other  reasons,  too,  besides  the  resisting  of 
stress  that  favor  its  use.* 


*  See  Richard's  "Manual  of  Machine  Construction. 


82  MACHINE  DESIGN. 

(1)  Its  appearance  is  far  finer,  giving  an  idea  of  complete- 
ness that  is  always  wanting  in  the  ribbed  frames. 

(2)  The  faces  of  a  box  frame  are  always  available  for  the 
attachment  of  auxiliary  parts  without  interfering  with  the  per- 
fection of  the  design. 

(3)  The  strength  can  always  be  increased  by  decreasing  the 
size  of  the  core,  without  changing  the  external  appearance  of 
the  frame,  and  therefore  without  any  work   whatever   on  the 
pattern  itself. 

The  cost  of  patterns  for  the  two  forms  is  probably  not  very- 
different,  the  pattern  itself  being  more  expensive  in  the  ribbed 
form,  and  the  necessary  core-boxes  adding  to  the  expense  in  the 
case  of  the  box  form.  The  expense  of  production  in  the  foundry, 
however,  is  greater  for  the  box  form  than  for  the  ribbed  form, 
because  core  work  is  more  expensive  than  "green-sand  "  work. 
The  balance  of  advantage  is  very  greatly  in  favor  of  box  forms, 
and  this  is  now  recognized  in  the  practice  of  the  best  designers 
of  machinery. 

To  illustrate  the  application  of  the  box  form  to  machine 
members,  let  the  table  of  a  planer  be  considered.  The  cross- 
section  is  almost  universally  of  the  form  shown  in  Fig.  51.  This 
is  evidently  a  form  that  would  yield  easily  to  a  force  tending  to 


FIG.  51. 

twist  it,  or  to  a  force  acting  in  a  vertical  plane  tending  to  bend  it. 
Such  forces  may  be  brought  upon  it  by  "strapping  down  work," 
or  by  the  support  of  heavy  pieces  upon  centers.  Thus  in  Fig.  52 
the  heavy  piece  E  is  supported  between  the  centers.  For  proper 
support  the  centers  need  to  be  screwed  in  with  a  considerable 
force.  This  causes  a  reaction  tending  to  separate  the  centers  and 
to  bend  the  table  between  C  and  D.  As  a  result  of  this  the  V's 


PROPORTIONS  OF  MACHINE  PARTS  AS  DICTATED  BY  STRESS.   83 

on  the  table  no  longer  have  a  bearing  throughout  the  entire 
surface  of  the  guides  on  the  bed,  but  only  touch  near  the  ends, 
the  pressure  is  concentrated  upon  small  surfaces,  the  lubricant  is 
squeezed  out,  the  V's  and  guides  are  "cut,"  and  the  planer  is 


FIG.  52. 


FIG.  53. 


FIG.  54. 


rendered  incapable  of  doing  accurate  work.  If  a  table  were  made 
of  the  box  form  shown  in  Fig.  53,  with  partitions  at  intervals 
throughout  its  length,  it  would  be  far  more  capable  of  maintain- 
ing its  accuracy  of  form  under  all  kinds  of  stress,  and  would  be 
more  satisfactory  for  the  purpose  for  which  it  is  designed.* 

The  bed  of  a  planer  is  usually  in  the  form  shown  in  section 
in  Fig.  54,  the  side  members  being  connected  by  "cross-girts" 
at  intervals.  This  is  evidently  not  the  best  form  to  resist  flexure 
and  torsion,  and  a  planer-bed  may  be  subjected  to  both,  either 
by  reason  of  improper  support  or  because  of  changes  in  the  form 
of  foundation.  If  the  bed  were  of  box  section  with  cross  parti- 
tions, it  would  sustain  greater  stress  without  undue  yielding. 
Holes  could  be  left  in  the  top  and  bottom  to  admit  of  supporting 
the  core  in  the  mold,  to  serve  for  the  removal  of  the  core  sand, 
and  to  render  accessible  the  gearing  and  other  mechanism  inside 
of  the  bed. 


*  Professor  Sweet  has  designed  and  constructed  such  a  table  for  a  large  mill- 
ing-machine. 


MACHINE  DESIGN. 


This  same  reasoning  applies  to  lathe-beds.  They  are  strained 
transversely  by  force  tending  to  separate  the  centers,  as  in  the 
case  of  "chucking";  torsionally  by  the  reaction  of  a  tool  cutting 
the  surface  of  a  piece  of  large  diameter;  and  both  torsion  and 
flexure  may  result,  as ,  in  the  case  of  the  planer-bed,  from  an 
improperly  designed  or  yielding  foundation.  The  box  form 
would  be  the  best  possible  form  for  a  lathe -bed;  some  diffi- 
culties in  adaptation,  however,  have  prevented  its  extended  use 
as  yet. 

These  examples  illustrate  principles  that  are  of  very  broad 
application  in  the  designing  of  machines. 

60.  Brackets. — Often  in  machines  there  is  a  part  that  pro- 
jects either  vertically  or  horizontally  and  sustains  a  transverse 


FIG.  55, 

stress;  it  is  a  cantilever,  in  fact.  If  only  transverse  stress  is  sus- 
tained, and  the  thickness  is  uniform,  the  outline  for  economy  of 
material  is  parabolic.  In  such  a  case,  however,  the  outline  curve 
of  the  member  should  start  from  the  point  of  application  of  the 
force,  and  not  from  the  extreme  end  of  the  member,  as  in  the 
latter  case  there  would  be  an  excess  of  material.  Thus  in  A, 
Fig-  55»  P  is  the  extreme  position  at  which  the  force  can  be  ap- 


PROPORTIONS  OF  MACHINE  PARTS  AS  DICTATED  BY  STRESS.  85 

plied.  The  parabolic  curve  a  is  drawn  from  the  point  of  appli- 
cation of  P.  The  end  of  the  member  is  supported  by  the  auxil- 
iary curve  c.  The  curve  b  drawn  from  the  end  gives  an  excess 
of  material.  The  curves  a  and  c  may  be  replaced  by  a  single 
continuous  curve  as  in  C,  or  a  tangent  may  be  drawn  to  a  at  its 
middle  point  as  in  B,  and  this  straight  line  used  for  the  outline, 
the  excess  of  material  being  slight  in  both  cases.  Most  of  the 
machine  members  of  this  kind,  however,  are  subjected  also  to 
other  stresses.  Thus  the  "housings"  of  planers  have  to  resist 
torsion  and  side  flexure.  They  are  very  often  supported  by  two 
members  of  parabolic  outline ;  and  to  insure  the  resistance  of  the 
torsion  and  side  flexure,  these  two  members  are  connected  at  their 
parabolic  edges  by  a  web  of  metal  that  really  converts  them  into 
-a  box  form.  Machine  members  of  this  kind  may  also  be  sup- 
ported by  a  brace,  as  in  D.  The  brace  is  a  compression  member 
and  may  be  stiffened  against  buckling  by  a  "web  "  as  shown,  or 
by  an  auxiliary  brace. 

61.  Other  Considerations  Governing  Form. — One  considera- 
tion governing  the  form  of  machine  parts  has  been  touched  upon 
in  the  preceding  sections.  It  may  be  well  to  state  it  here  as  a 
general  principle:  Other  considerations  being  equal  the  form  of 
a  member  should  be  that  which  can  be  most  cheaply  produced 
both  as  regards  economy  of  material  and  labor. 

Another  element  enters  into  the  form  of  cast  members.  Cast- 
ings, unless  of  the  most  simple  form,  are  almost  invariably  sub- 
jected to  indeterminate  shrinkage  stresses.  Some  of  these  are 
undoubtedly  due  to  faulty  work  on  the  part  of  the  molder,  others 
are  induced  by  the  very  form  which  is  given  the  piece  by  the  de- 
signer. They  cannot  be  eliminated  entirely,  but  the  danger  can 
be  minimized  by  paying  attention  to  these  general  laws: 

(a)  Avoid  all  sharp  corners  and  re-entrant  angles. 

(b)  All  parts  of  all  cross-sections  of  the  member  should  be 
as  nearly  of  the  same  thickness  as  possible. 


86  MACHINE  DESIGN. 

(c)  If  it  is  necessary  to  have  thick  and  thin  parts  in  the  same 
casting,  the  change  of  form  from  one  to  the  other  should  be  as 
gradual  as  possible. 

(d}  Castings  should  be  made  as  thin  as  is  consistent  with  con- 
siderations of  strength,  stiffness,  and  resistance  to  vibration. 


CHAPTER  VII. 


RIVETED    JOINTS. 

62.  Methods  of  Riveting.  —  A  rivet  is  a  fastening  used  to  unite 
metal  plates  or  rolled  structural  forms,  as  in  boilers,  tanks,  built- 
up  machine  frames,  etc.  It  consists  of  a  head,  A,  Fig.  56,  and  a 
straight  shank,  B.  It  is  inserted,  usually  red-hot,  into  holes, 
either  drilled  or  punched  in  the  parts  to  be  connected,  and  the 
projecting  end  of  the  shank  is  then  formed  into  a  head  (see  dotted 
lines)  either  by  hand-  or  machine-riveting.  A  rivet  is  a  permanent 
fastening  and  can  only  be  removed  by  cutting  off  the  head.  A 
row  of  rivets  joining  two  members  is  called  a  RIVETED  JOINT  or 
SEAM  OF  RIVETS.  In  hand-riveting  the  projecting  end  of  the 
shank  is  struck  a  quick  succession  of  blows  with  hand  hammers 
and  formed  into  a  head  by  the  workman.  A  helper  holds  a  sledge 


FIG.  56. 


FIG.  58. 


or  "dolly  bar"  against  the  head  of  the  rivet.  In  "button-set" 
or  "snap"  riveting,  the  rivet  is  struck  a  few  heavy  blows  with  a 
sledge  to  "upset"  it.  Then  a  die  or  "button  set,"  Fig.  57,  is 
held  with  the  spherical  depression,  B,  upon  the  rivet ;  the  head, 
A,  is  struck  with  the  sledge,  and  the  rivet  head  is  thus  formed. 
In  machine-riveting  a  die  similar  to  B  is  held  firmly  in  the  ma- 
chine and  a  similar  die  opposite  to  it  is  attached  to  the  piston  of 

87 


MACHINE  DESIGN. 

a  steam,  hydraulic,  or  pneumatic  cylinder.  A  rivet,  properly 
placed  in  holes  in  the  members  to  be  connected,  is  put  between 
the  dies  and  pressure  is  applied  to  the  piston.  The  movable  die 
is  forced  forward  and  a  head  formed  on  the  rivet. 

The  relative  merits  of  machine-  and  hand-riveting  have  been 
much  discussed.  Either  method  carefully  carried  out  will  pro- 
duce a  good  serviceable  joint.  If  in  hand-riveting  the  first  few 
blows  be  light  the  rivet  will  not  be  properly  upset,  the  shank  will 
be  loose  in  the  hole,  and  a  leaky  rivet  results.  If  in  machine- 
riveting  the  axis  of  the  rivet  does  not  coincide  with  the  axis  of 
the  dies,  an  off -set  head  results.  (See  Fig.  58.)  In  large  shops 
where  work  must  be  turned  out  economically  in  large  quantities, 
machines  must  be  used.  But  there  are  always  places  inacces- 
sible to  machines,  where  the  rivets  must  be  driven  by  hand.* 

63.  Perforation  of  Plates. — Holes  for  the  reception  of  rivets 
are  usually  punched,  although  for  thick  plates  and  very  careful 
work  they  are  sometimes  drilled.  If  a  row  of  holes  be  punched 
in  a  plate,  and  a  similar  row  as  to  size  and  spacing  be  drilled  in 
the  same  pkte,  testing  to  rupture  will  show  that  the  punched  plate 
is  weaker  than  the  drilled  one.  If  the  punched  plate  had  been 
annealed  it  would  have  been  nearly  restored  to  the  strength  of 
the  drilled  one.  If  the  holes  had  been  punched  ^  inch  to  |  inch 
small  in  diameter  and  reamed  to  size,  the  pkte  would  have  been 
as  strong  as  the  drilled  one.  These  facts,  which  have  been  ex- 
perimentally determined,  point  to  the  following  conclusions: 
First,  punching  injures  the  material  and  produces  weakness. 
Second,  the  injury  is  due  to  stresses  caused  by  the  severe  action 
of  the  punch,  since  annealing,  which  furnishes  opportunity  for 
equalizing  of  stress,  restores  the  strength.  Third,  the  injury 
is  only  in  the  immediate  vicinity  of  the  punched  hole,  since  ream- 
ing out  Jj-  inch  or  less  on  a  side  removes  all  the  injured  material. 

*  See  Sec.  75  for  discussion  of  the  importance  of  holding  rivet  under  presume 
until  it  is  cooled;  and  the  advantage  of  large  rivets  over  small. 


RIVETED  JOINTS. 


In  ordinary  boiler  work  the  plates  are  simply  punched  and  riveted. 
If  better  work  is  required  the  plates  must  be  drilled,  or  punched 
small  and  reamed,  or  punched  and  annealed.  Drilling  is  slow 
and  therefore  expensive;  annealing  is  apt  to  change  the  plates 
and  requires  large  expensive  furnaces.  Punching  small  and  ream- 
ing is  probably  the  best  method.  In  this  connection,  Prof.  A. 
B.  W.  Kennedy  (see  Proc.  Inst.  M.  E.,  1888,  pp.  546-547)  has 
called  attention  to  the  phenomenon  of  greater  unit  tensile  strength 
of  the  plate  along  the  perforations  than  of  the  original  unperforated 
plate.*  Stoney  ("  Strength  and  Proportion  of  Riveted  Joints," 
London,  1885)  has  compiled  the  following  table: 

TABLE  I. — RELATIVE  PERCENTAGE  OF  STRENGTH  OF  STEEL  PLATES  PERFORATED 
IN  DIFFERENT  WAYS. 


Specimens. 

Unit  Strength  of  Net  Section  between  Holes  compared 
with  that  of  the  Solid  Plate  (100  Per  Cent). 

i  Inch. 

J  Inch. 

1  Inch. 

i  Inch, 

Punched  
Punched  and  annealed  
Drilled  

Per  Cent. 

IOI  .O 

105.6 
113.8 

Per  Cent 
94.2 
105.6 
III.  I 

Per  Cent. 
82.5 

IOI  .O 

106.4 

Per  Cent. 
75-8 
100.3 
106.  i 

For  punched  and  reamed  holes  the  .same  percentages  may 
be  used  as  for  drilled. 

Professor  Kennedy  gives  constants  which  may  be  obtained 
from  the  following  formula : 

Excess  of  unit  strength  of  drilled  steel  plates  in  net  section 
over  unperforated  section 


/  is  the  thickness  of  plate  in  inches  and  r  the  ratio  of    pitch 

*  This  reported  phenomenon  is  corroborated  by  tests  made  at  Watertown 
Arsenal.  See  Tests  of  Metals,  1886,  pp.  1264,  1557.  It  is  fully  explained  by 
the  condition  of  localized  stress  and  the  consequent  prevention  of  lateral 
contraction. 


go 


MACHINE  DESIGN. 


divided  by  diameter  of  hole.      No  data  exist  relative  to   iron 
plates  in  this  matter. 

64.  Kinds  of  Joints. — Riveted  joints  are  of  two  general  kinds : 
First,  LAP-JOINTS,  in  which  the  sheets  to  be  joined  are  lapped  on 
each  other  and  joined  by  a  seam  of  rivets,  as  in  Fig.  $ga.  Second, 
BUTT-JOINTS,  in  which  the  edges  of  the  sheets  abut  against  each 

FIG.  59. 


I 

oil 

1 

oil  1 

OOji 

ool 

0° 
0! 

o  o 
o  o 

p 

i'l 

lo 

OiO 
O    i    O 

o°Po 

OIO 

c 

i 

C 

1 

[I 
ll 
II 
II 
ll 

i 

FIG.  60. 


other,  and  a  strip  called  a  "cover-plate"  or  "butt-strap"  is  riv- 
eted to  the  edge  of  each  sheet,  as  in  c.     In  recent  years  a  lap-joint 


RIVETED  JOINTS.  91 

with  a  single  cover-plate  has  been  used  somewhat.  It  is  shown 
in  Fig.  590. 

There  are  two  chief  kinds  of  riveting:  Single,  in  which  there 
is  but  one  row  of  rivets,  as  in  Fig.  59**;  and  double,  where  there 
are  two  rows. 

Double  riveting  is  subdivided  into  " chain-riveting,"  Fig.  596, 
and  "zigzag  "  or  "staggered  "  riveting,  Fig.  $gd. 

Lap-joints  may  be  single,  double  chain,  or  double  staggered 
riveted. 

Butt-joints  may  have  a  single  strap  as  in  c,  or  double  strap; 
i.e.,  an  exactly  similar  one  is  placed  on  the  other  side  of  the  joint. 
Butt-joints  with  either  single  or  double  strap  may  be  single, 
double  chain,  or  double  staggered  riveted.  In  butt-joints  single 
cover-plates  should  be  nine-eighths  of  the  thickness  of  plates 
and  double  cover-plates  five-eighths. 

To  sum  up,  there  are: 

f  Single-riveted 

Lap-joints \  Double  chain-riveted 

(  Double  staggered-riveted 


Butt-joints. 


f  Single-riveted 
Single-strap j  Double  chain-riveted 

[  Double  staggered-riveted 

f  Single-riveted 
Double-strap j  Double  chain-riveted 

[  Double  staggered-riveted 


The  demands  of  modern  practice  have  added  triple  and  quad- 
ruple joints  to  the  foregoing.  In  high-pressure  cylindrical  boilers, 
for  instance,  common  practice  is  to  employ  for  the  longitudinal 
seam  the  highly  efficient  joint  shown  in  Fig.  60.  Here  we  have 
a  triple- riveted  butt-joint  with  double  cover-plates;  on  each  side 
of  the  joint  two  rows  of  rivets  are  in  double  shear  and  one  row, 
the  outer,  is  in  single  shear. 

65.  Failure  of  Joint. — A  riveted  joint  may  yield  in  any  one  of 
four  ways:  First,  by  the  rivet  shearing  (Fig.  6ia);  second,  by  the 
plate  yielding  to  tension  on  the  line  AB  (Fig.  616);  third,  by  the 


92  MACHINE  DESIGN. 

rivet  tearing  out  through  the  margin,  as  in  c;  fourth,  the  rivet 
and  sheet  bear  upon  each  other  at  D  and  E  in  d,  and  are  both 
in  compression.  If  the  unit  stress  upon  these  surfaces  becomes 
too  great,  the  rivet  is  weakened  to  resist  shearing,  or  the  plate 
to  resist  tension,  and  failure  may  occur.  This  pressure  of  the 


FIG.  61. 

rivet  on  the  sheet  is  called  "bearing  pressure."  It  is  obvious 
that  the  strongest  or  most  efficient  joint  in  any  case  will  be  one 
which  is  so  proportioned  that  the  tendency  to  fail  will  be  equal 
in  all  of  the  ways. 

66.  Strength  of  Materials  Used. — As  a  preliminary  to  the 
designing  of  joints  it  is  necessary  to  know  the  strength  of  the 
rivets  to  resist  shear,  of  the  plate  to  resist  tension,  and  of  the 
rivets  and  plates  to  resist  bearing  pressure.  These  values  must 
not  be  taken  from  tables  of  the  strength  of  the  materials  of  which 
the  plate  and  rivets  are  made,  but  must  be  derived  from  experi- 
ments upon  actual  riveted  joints  tested  to  rupture.  The  reason 
for  this  is  that  the  conditions  of  stress  are  modified  somewhat  in 
the  joint.  For  instance,  in  single-strap  butt-joints,  and  in  lap- 
joints,  the  line  of  stress  being  the  center  line  of  plates,  and  the 
plates  joined  being  offset,  flexure  results  and  the  plate  is  weaker 
to  resist  tension,  the  rivets  in  the  mean  time  being  subjected  to 
tension  as  well  as  shear;  if  the  joint  yield  to  this  stress  in  the 
slightest  degree  the  "bearing  pressure  "  is  localized  and  becomes 
more  destructive.  The  effect  of  friction  between  the  surfaces 
of  the  plates  under  the  pressure  at  which  they  are  "gripped  " 
by  the  rivets  is  another  item  of  considerable  importance.  Ex- 


RIYETED  JOINTS. 


93 


tensive  and  accurate  experiments  have  been  made  upon  actual 
joints  and  the  results  have  been  published.* 

The  following  table  has  been  compiled  as  representing  fair 
average  results,  and  the  values  here  given  may  be  used  for  ordi- 
nary joints: 

TABLE  II. — VALUES  OF  jt,  /«,  AND  fc  FOR  DIFFERENT  KINDS  OF  JOINTS. 


Kind  of  Joint. 

Iron. 

Steel. 

ft 

f, 

fc 

// 

h 

fc 

Lap-joint,    single-riveted,     punched 

holes  

40000 

38000 

67000 

475°° 

85000 

Lap-joint,  single-riveted,  drilled  holes. 

45000 

36000 

67000 

45000 

85000 

Lap-joint,   double-riveted,   punched 

holes  

45000 

40000 

67000 

48000 

85000 

Lap-joint,  double-riveted,  drilled  holes 

50000 

38000 

67000 

46000 

85000 

Butt-joints,  single  cover  :   Use  values 

given  for  lap-joints. 

Butt-joints,  double  cover,  single-riv- 

/.' 

fc 

u 

fc 

eted,  punched  holes  

40000 

42500 

89000 

48000 

100000 

Butt-joints,  double  cover,  single-riv- 

eted, drilled  holes  

45000 

41000 

89000 

46000 

100000 

Butt-joints,  double  cover,  double-riv- 

eted, punched  holes  

45000 

38000 

89000 

47500 

1  00000 

Butt-joints,  double  cover,  double-riv- 

eted, drilled  holes  

50000 

36000 

89000 

45000 

lOOOOO 

t  Original  plate  

50000 

60000 

t  Original  bar  

45000 

52000 

The  Master  Steam  Boiler-Makers'  Assn.,  as  the  result  of 
tests  conducted  by  its  committee,  recommends,  for  iron  rivets, 
/s  =  42000  and// -=40000;  for  steel  rivets,  fs=  46000,  //  =  44000. 

It  will  be  noted  that  the  values  of  //  are  not  given  for  steel. 
The  tensile  strength  of  steel  varies  through  a  considerable  range 
due  largely  to  differences  in  chemical  constitution;  it  also  follows 
a  rough  law  of  inverse  proportion  to  the  thickness  of  plates; 
i.e.,  thin  plates  will  be  almost  sure  to  show  higher  tensile  strength 

*  See  Proc.  Inst.  of  Mech.  Eng.,  1881,  1882,  1885,  1888;  Tests  of  Metals, 
Watertown  Arsenal,  1885,  1886,  1887,  1891,  1895,  1896.  Stoney's  "  Strength  and 
Proportions  of  Riveted  Joints,"  London,  1885. 

t  If  the  original  material  varies  from  this,  the  values  given  above  should  be 
varied  proportionately. 


94  MACHINE  DESIGN. 

than  thicker  plates  of  the  same  composition.  Furthermore, 
the  method  of  perforation  greatly  affects  the  strength  of  the  plates, 
as  has  been  pointed  out  in  §  63.  Ordinary  boiler-plates  have  a 
unit  tensile  strength  ranging  from  55,000  Ibs.  to  62,000  Ibs.  per 
square  inch.  For  ordinary  calculations  }t  may  be  taken  as 
55,000  Ibs.  for  punched  plates  and  60,000  for  drilled  plates. 
The  shearing  strength  of  rivets  also  varies  inversely  as  their  size, 
but  these  differences  are  slight. 

67.  Strength,  Proportions,  and  Efficiency  of  Joints. — No 
riveted  joint  can  be  as  strong  as  the  unperforated  plate.  The 
ratio  of  strength  of  joint  to  strength  of  unperforated  plate  is 
called  the  JOINT  EFFICIENCY. 

As  stated  in  §  65  the  highest  efficiency  for  a  joint  is  obtained 
when  the  relations  between  thickness  of  plate,  diameter  of  rivet, 
pitch,  and  margin  are  such  that  the  tendency  for  the  joint  to 
fail  in  any  one  way  does  not  exceed  the  tendency  for  it  to  fail 
in  any  other  way.  Formulae  can  be  developed  for  finding  their 
proper  values  for  each  form  of  joint. 

Let  d  =  diameter  of  rivet-hole  in  inches ; 
a  =  pitch  of  rivets  in  inches; 
/  =  thickness  of  plates  in  inches ; 

ft  =  tensile  strength  of  plates  in  pounds  per  square  inch; 
}e  =  crushing  strength  of  rivets  or  plates,  if  rivets  are  in 

single  shear,  pounds  per  square  inch ; 
//=  crushing  strength  of  rivets  or  plates,  if  rivets  are  in 

double  shear,  pounds  per  square  inch; 
/.=  shearing  strength  of  rivets  in  single  shear,  pounds  per 

square  inch; 
//=  shearing  strength  of    rivets    in    double  shear,  pounds 

per  square  inch. 

Each  joint  may  be  treated  as  if  made  up  of  a  successive  series 
of  similar  strips,  each  unit  strip  having  a  width  equal  to  a,  the 
distance  between  centers  of  two  consecutive  rivets  in  the  same 


RIVETED  JOINTS. 


95 


row  (see  Fig.  62).  If  the  stresses  and  proportions  for  one  such 
strip  are  determined,  the  results  obtained  will,  of  course,  apply 
to  all  of  the  others,  and  consequently  to  the  whole  joint.  Con- 
sider such  a  strip  of  thickness  /  and  width  a. 


O 
O 


FIG.  62. 

Let  P=  ultimate  tensile  strength  of  unperforated  strip,  pounds; 
T  =  ultimate  tensile  strength  of  net  section  of  strip,  pounds; 
S=  ultimate  shearing  resistance  of  all  rivets  in  strip, 

pounds ; 
C=  ultimate  crushing  resistance  of  all  rivets  or  sides  of 

holes,  pounds; 
E  =  efficiency  of  joint. 

To  illustrate  this  method,  consider  first  the  simplest  joint, 
i.e.,  the  single-riveted  lap-joint. 

The  unperforated  strip  has  a  tensile  strength 


P=atft 


(i) 


Along  the  row  of  rivets  the  net  width  of  plate  is  less  than  the 
total  width  of  the  strip  by  an  amount  equal  to  the  diameter  of 
the  rivet,  and  consequently  the  net  tensile  strength 'of  the  strip 
\s  expressed  by  the  equation 

T-(a-$tft (2) 

In  each  unit  strip  there  is  but  a  single  rivet  with  but  one  sur- 
face in  shear,  hence 


(3) 


(,6  MACHINE  DESIGN. 

The  crushing  resistance  of  the  rivet,  or  of  the  plate  around 
the  hole,  may  be  written  as 

C  =  dtfe  .........     (4) 

For  highest  efficiency  T  =  S  =  C. 

Equating  5  and  C,  (3)  and  (4),  .>j8$4d2ft=dtfc. 


(5) 


This  equation  gives  the  proper  theoretical  value  of  d  for  a 
given  value  of  /,  and  for  materials  represented  by  fc  and  }8. 
Equating  T  and  S,  (2)  and  (3), 


(6) 


This  gives  the  proper  theoretical  pitch.  The  efficiency  of  the 
joint  is  obtained  by  dividing  T,  S,  or  C  by  P. 

In  most  cases  the  values  of  d  and  a  as  determined  by  (5) 
and  (6)  cannot  be  strictly  adhered  to.  Stock  sizes  of  rivets 
must  be  used  in  practice,  and  there  are  also  limitations  connected 
with  the  largest  sizes  it  is  convenient  to  drive.  These  equations, 
furthermore,  do  not  take  into  consideration  the  stresses  set  up  in 
the  rivets  when  their  shrinkage,  due  to  cooling,  is  resisted  by 
the  plates,  an  item  which  may  become  excessive  with  the  smaller 
diameters.  The  spacing  of  the  rivets  must  also  be  modified  quite 
frequently  by  the  proportions  of  the  parts  to  be  connected,  by 
allowance  for  proper  space  to  form  the  heads,  and  by  provision 
for  tightness.  In  practice  it  is  therefore  often  necessary  to  depart 
from  these  values. 

It  must  be  borne  in  mind,  however,  that  any  departure  from 
the  values  of  d  and  a  given  in  (5)  and  (6)  destroys  the  equality 


R1YETED  JOINTS.  97 

between  T,  S,  and  C,  and  if  such  departure  is  made,  the  actual 
value  of  T,  S,  and  C  should  be  determined  (by  substitution  of 
the  values  of  d  and  a  decided  upon).  The  efficiency  of  the 
joint  will  then  be  found  by  dividing  whichever  has  the  smallest 
value,  T,  S,  or  C,  by  P. 

If  the  REAL  efficiency  of  the  joint  is  desired,  the  value  of 
T  must  be  obtained  by  increasing  }t  by  the  amount  called  for  by 
the  perforation  of  the  plates.  As  explained  in  §  63  this  will  be 

/  2  +  —7=  j  (    '         )  per  cent  greater  than  the  /<  of  the  original,  un- 

perforated  plate. 

68.  Problem. — The  following  problem  illustrates  the  method 
of  using  Table  II  in  connection  with  the  formulae  (5)  and  (6). 

What  should  be  the  dimensions  of  rivet-hole  and  pitch  for  a 
single-riveted  lap-joint  for  f-inch  iron  plates  using  iron  rivets? 

Table  II  gives  as  values  of  }t,  j,,  and  fc  40,000,  38,000,  and 
67,000  Ibs.  per  square  inch,  respectively,  for  this  fortn  of  joint. 

<-t"-.375". 

Substituting  these  values,  equation  (5)  becomes 

'-"7X.375X-Hl-.84" 

and  equation  (6) 

.  7854  X.8~42X  38000 
0  =  -^-^  —+.84  =  2.24  inches. 

.375X40000 

69.  Proportions     of    Single-riveted    Lap-joints. — Table    III 
and  Table  IV  have  been  computed  in  this  way.     As  Table  IV 
refers  to  steel  joints,  the  values  of  /,,  }s,  and  fe  are  55,000,  47,500, 
and  85,000  Ibs.  per  square  inch,  respectively. 


MACHINE  DESIGN. 


TABLE  III. — PROPORTIONS  OF  SINGLE-RIVETED  LAP-JOINTS,  IRON  PLATES,  AND- 
RIVETS,  PUNCHED  HOLES. 


/ 

fc 

.  7854^/5 

d 

a 

I>27'77 

<fc 

A 

.42 

.  12 

\ 

•56 

•49 

t 

.70 
.84 

.86 
.24 

f 

t. 

\ 

.12 

.98 

f~rt 

2-2^ 

1 

.40 

3-73 

|-i 

2-2f 

1 

.68 

4-47 

Mj 

1 

.96 

5.22 

2^-3 

I 

.24 

5-96 

I-il 

2|-2f 

I* 

•52 

6.71 

li-tf 

2f-3 

Column  i  gives  the  thickness  of  plate;  columns  2  and  3  give 
the  corresponding  calculated  values  of  d  and  a  for  joint  of  maxi- 
mum efficiency;  columns  4  and  5  give  the  values  of  d  and  a 
as  compiled  by  Twiddell  in  the  Proc.  Inst.  of  M.  E.,  1881,  pp. 
293-295,  from  boiler-makers'  practice.  It  will  be  noted  that  the 
rivets  used  in  practice  (see  column  4)  are  considerably  smaller 
in  diameter  than  those  called  for  in  column  2,  and  that  this  differ- 
ence grows  more  and  more  marked  as  the  thickness  of  the  plate 
increases.  The  reason  for  this  is  that  the  difficulty  in  driving 
rivets  increases  very  rapidly  with  their  size,  if-  or  if  inches 
being  the  largest  rivet  that  can  be  driven  conveniently.  The 
equality  of  strength  to  resist  bearing  pressure  and  shear  is  there- 
fore sacrificed  to  convenience  in  manipulation.  As  the  diameter 
of  the  rivet  is  increased  the  area  to  resist  bearing  pressure  in- 
creases less  rapidly  than  the  area  to  resist  shear  (the  thickness 
of  the  plate  remaining  the  same),  the  former  varying  as  d  and  the 
latter  as  d2;  therefore  if  d  is  not  increased  as  much  as  is  neces- 
sary for  equality  of  strength,  the  excess  of  strength  will  be  to  re- 
sist bearing  pressure.  If  the  other  parts  of  the  joint  are  made 
as  strong  as  the  rivet  in  shear,  and  this  strength  is  calculated 
from  the  stress  to  be  resisted,  the  joint  will  evidently  be  correctly 
proportioned.  As  machine-riveting  comes  into  more  general 


RIYETED  IOINTS. 


99 


use  and  pneumatic  tools  are  used  in   "hand-work,"   this  dis- 
crepancy will  tend  to  disappear. 

TABLE  IV.— PROPORTIONS  OF  SINGLE-RIVETED  LAP-JOINTS,  STEEL  PLATES,  AND 
RIVETS,  PUNCHED  HOLES. 


fr 

.»*S4d% 

t 

'f< 

d 

a 

A 

•43 

1.  08 

•47 

l| 

i 

•57 

i-45 

.61 

JA 

t^ 

1.  81 

.81 

2 

1 

.86 

2.17 

•94 

2A 

* 

1.14 

2.89 

1.19 

3 

Column  i  gives  the  thickness  of  the  plate;  columns  2  and  3 
give  the  values  of  d  and  a  calculated  for  joint  of  maximum  effi- 
ciency; columns  4  and  5  give  proportions  from  practice,  the 
authority  being  Moberly  (see  Stoney,  "  Strength  and  Proportions 
of  Riveted  Joints,"  p.  80).  It  will  be  noted  how  closely  the 
theory  and  practice  agree  here  for  boiler  joints.* 

70.  Single-riveted  Butt-joints.  —  To  develop  the  general 
formulae  for  the  values  of  a  and  d  for  single -riveted  butt-joints 
with  double  cover-plates  the  same  general  method  used  in  §  67 
applies. 

In  this  case  the  rivets  are  in  double  shear.     Therefore 


while  T  =  (a-d)tft,  (2),  as  before  and 
C=dt}c'. 
Equating  5  and  C,  (7)  and  (8), 


(7) 


(8) 


*  For  further  data,  compiled  from  American  practice,  see  sec.  73. 


ioo  MACHINE  DESIGN. 


/.'-*/.';    .'.       /.'-*' 
and  d  =  .64fy  .........     (9) 

Equating  T  and  5,  (2}  and  (7), 


For  Double-riveted  Lap-joints  the  unit  strip  contains  two  rivets, 
each  in  single  shear.    The  following  equations  cover  the  case  : 


2-j/. 

2dt}c, 

(H) 


71.  Double-riveted  Butt-joint.  —  For  double  riveted  butt- 
joints,  double  cover-plate,  either  chain  or  staggered  riveting,  there 
are  two  rivets  in  double  shear  for  each  unit  strip. 

T  =  (a-d)tft, 


RIVETED  JOINTS.  lor 

(13) 

(14) 

72.  General  Formulae.  —  The  following  general  equations  for 
riveted  joints  have  been  developed  by  Mr.  W.  N.  Barnard:* 

The  unit  strip  is  of  width  equal  to  the  pitch,  the  maximum 
pitch  being  taken  unless  all  rows  have  the  same  pitch. 

The  general  expression  for  the  net  tensile  strength  of  the  unit 
strip  is 

T  =  (a-d}tft  ........     (15) 

The  general  expression  for  resistance  to  shearing  of  the  rivets 
in  the  unit  strip  is 

*  mid2        2mxd2 

S=—fs  +  --  //,      .....     (16) 

4  4 

in  which  n  equals  the  number  of   rivets  in  single  shear  and  m 
equals  the  number  of  rivets  in  double  shear. 

The  general  expression  for  resistance  to  crushing  of  the  unit 
strip  is 

C=ndtfc+mdt}c'  ........     (17) 

The  tensile  resistance  9f  the  solid  strip  is 

P  =  atft  ...........     (18) 

Equating  S  and  C,  (16)  and  (17),  and  transposing,  we  get 


*  See  article,  "  General  Formulas  for  Efficiency  and  Proportions  for  Riveted 
Joints,"  by  Professor  J.  H.  Barr  in  Sibley  Journal  of  Engineering,  Oct.,  1900. 


102  MACHINE  DESIGN. 

Equating  Tand  C,  (15)  and  (17), 
Or,  equating  T  and  5,  (15)  and  (16), 


The  following  equation   for  efficiency  has    been    developed 
on  the  assumption  that  T=S  =  C. 

g 
From  E=~B  we  get,  by  substitution  and  transposition, 


This  equation  is  useful  in  finding  the  limiting  efficiency  ci 
joint  for  any  form  and  materials;  the  actual  proportions  adopted 
may  give  a  lower  efficiency,  but  can  never  give  a  higher  efficiency.* 
73.  Proportions  of  Joints.  —  In  American  practice  it  will  be 
found  that  there  is  more  or  less  departure  from  the  proportions 
which  would  be  arrived  at  by  the  strict  application  of  the  prin- 
ciples laid  down  in  the  preceding  articles.  This  variation  is 
due  to  several  considerations.  Chief  among  them  is  the  practical 
difficulty  of  driving  large  rivets,  thus  leading  to  the  adoption  of 
rivet  diameters  with  reference  to  convenience  of  manipulation 
rather  than  efficiency  of  joint.  As  machines  displace  handwork 
the  reason  for  this  departure  disappears  and  there  is  an  increasing 
tendency  to  use  the  larger  and  more  correct  rivet  diameters. 
Conservatism  must  be  reckoned  with  here  and  also  in  the  failure 
to  recognize  the  fact  that  rivet  diameters  do  not  depend  solely 

*  In  the  Proceedings  of  the  Inst.  of  M.  E.,  1881,  there  is  an  article  entitled 
"On  Riveting,  with  Special  Reference  to  Ship-work,"  M  Le  Baron  Clauzel, 
which  enters  deeply  into  the  development  of  general  formulae. 


RIVETED  JOINTS.  103 

upon  the  thickness  of  plates,  but  also  should  vary  with  the  kind 
of  joint.  Practice  tends  to  hold  to  one  diameter  of  rivet  for  each 
thickness  of  plate,  irrespective  of  the  kind  of  joint. 

Another  item  of  practical  importance  is  tightness  against 
leakage  under  pressure.  Most  formulae  are  developed  without 
consideration  of  this  important  factor.  From  a  practical  point 
of  view,  the  joint  fails  when  it  begins  to  leak,  actual  rupture 
need  not  take  place.  The  topic  of  the  allowable  maximum 
pitch  as  governed  by  experience  with  tightness  of  joints  is 
•discussed  in  §  80. 

The  margin  in  a  riveted  joint  is  the  distance  from  the  edge  of 
the  sheet  to  the  rivet  hole.  This  must  be  made  of  such  value  that 
there  shall  be  safety  against  failure  by  the  rivet  tearing  out. 
There  can  be  no  satisfactory  theoretical  determination  of  this 
value;  until  recently  it  has  been  held  that  practice  and  experi- 
ments with  actual  joints  showed  that  a  joint  would  not  yield  in 
this  way  if  the  margin  were  made  =  d  =  diameter  of  the  rivet 
hole.  This  is  a  safe  rule  for  iron  rivets  in  steel  plates  for  any 
type  of  joint.  Where  steel  rivets  are  used  it  will  be  well  to 

Increase  this  to  -  d.     (See  Power,  Aug.,  1905.) 

4 

The  distance  between  the  center  lines  of  rows  may  be  taken 
not  less  than  2.5^  for  double-chain  riveting,  and  1.88^  for 
double-staggered  riveting.  This  will  insure  safety  against 
zigzag  tearing  of  the  plate,  but  brings  the  heads  very  close  together. 
From  these  values  and  those  of  margins,  as  just  discussed,  the 
proper  amount  of  lap  can  readily  be  determined  for  any  kind  of 
joint. 

74.  Relative  Efficiencies  of  Various  Kinds  of  Joints. — The 
.actual  efficiencies  of  joints  when  tested  show  some  departure 
from  the  calculated  ideal  efficiencies.  The  following  Table  (V) 
lias  been  compiled  from  the  results  of  tests  to  show  roughly  the 
relative  efficiencies  of  various  types  of  joints: 


104 


MACHINE  DESIGN. 


TABLE  V. — RELATIVE  EFFICIENCY  OF  IRON  JOINTS. 


Efficiency 
Per  Cent. 


Original  solid  plate 

Lap-joint,  single-riveted,  punched 

drilled 

' '          double     "      

Butt-joint,  single  cover,  single-riveted 

"         "       double-riveted 

' '  double    ' '       single-riveted 

"  "        "       double-riveted 


100 

45 

50 


45-5° 
60 

55 
66 


RELATIVE  EFFICIENCY  OF  STEEL  JOINTS. 


Efficiency  Per  Cent. 


Thic 
i-l 

kness  of  PI 
*-i 

ates. 
*-J 

Original  solid  plate  

IOO 

IOO 

IOO 

Lap-joint,  single-riveted,  punched  
drilled  

5° 
55 

45 

5° 

40 

45 

7> 

7° 

6; 

drilled  

80 

75 

70 

Butt-joint,  double  cover,  single-riveted,  drilled  
"          "       double-riveted,  punched.  .. 

70 

75 

65 

70 

60 
65 

drilled  

80 

75 

70 

These  tables  are  from  Stoney's  "Strength  and  Proportions  of  Riveted  Joints." 

Triple  and  quadruple  riveted  butt-joints  with  double  cover- 
plates  show  efficiencies  ranging  from  80  to  90  per  cent.* 

75.  Slippage. — At  about  25  to  35  per  cent  of  its  ultimate  load 
SLIPPAGE  takes  place  in  a  riveted  joint.  This  is  probably  due 
to  the  fact  that  at  this  load  the  friction  between  the  plates,  owing 
to  the  pressure  exerted  on  them  by  the  rivets,  is  overcome.  It 
has  been  found  the  larger  the  cross-sectional  area  of  the  rivet 
the  greater  the  percentage  of  ultimate  load  which  can  be  with- 
stood without  slippage.  It  has  also  been  found  that  large  rivet- 
heads  are  better  than  small  ones  for  the  same  reason. 


*  For  details  of  joints  tested,  see  Tests  of  Metals,  Watertown  Arsenal,  1806- 


RIVETED  JOINTS.  105 

The  importance  of  the  consideration  of  slippage  has  been  fully 
established  by  the  work  of  Professor  Bach  ("  Die  Maschinen- 
elemente,"  Qth  ed.  pp.  164-195).  His  careful  and  exhaustive 
experiments  prove  that: 

1.  In  cooling    the  rivet  shrinks  away  from  the  walls  of  the 
hole. 

2.  In  consequence  of  this,  there  is  no  tendency  to  shear  off 
the  rivet  until  after  the  joint  has  failed,  for  all  practical  purposes, 
by  losing  tightness  because  of  slippage. 

3.  The  percentage  of  the  ultimate  or  rupture  load  at  which 
slippage  takes  place  varies  according  to  three  items  : 

a.  It  is  directly  proportional  to  the  square  of  the  diameter 

of  the  rivet.     From  this  the  desirability  of  using  large 
rivets  rather  than  small  is  further  established. 

b.  It  is  increased  by  calking,  especially  if  both  rivet  heads 

are  calked  as  well  as  the  plate  edges. 

c.  It  is  greatly  increased  by  holding  the  rivets  under  maxi- 

mum pressure  until  they  are  cool  enough  to  have  set. 

This  gives   better   results   than   blows,  light   pressure, 

or  early  removal  of  pressure. 

Professor  Bach  argues  that  joints  should  not  be  proportioned 
with  reference  to  the  ultimate  or  rupture  strength.  He  claims 
that  the  maximum  pitch  is  determined  by  the  condition  of  tight- 
ness against  springing  open  between  rivets  when  pressure  is 
applied.  The  minimum  pitch  is  that  fixed  by  the  spacing 
of  rivet  heads  which  is  the  least  which  will  permit  calking  them. 
Between  these  limits  he  chooses  pitch: 

1.  So  that  the  safe  resistance  to  slippage  (as  experimentally 
determined  by  him)  is  equated  to  the  stress  on  the  joint  due  to 
the  diameter  of  the  vessel  and  the  pressure. 

2.  So  that  the  unit  stress  in  the  plate  between  the  rivets  shall 
not  exceed  the  safe  working  value  of  the  plate  material  when  the 
strength  of  the  perforated  section  is  equated  to  the  stress  due  to 
the  diameter  and  pressure. 


io6 


MACHINE  DESIGN. 


Plate  I  shows  graphically  the  proportions  of  riveted  joints  as 
determined  from  Professor  Bach's  formulas  by  M.  Shibata  in 
the  American  Machinist,  Vols.  26  and  27. 

76.  Rivet  Size  and  Proportions. — In  general  the  rivet  should 
have  a  shank  A  inch  smaller  in  diameter  than  the  hole  to  be 
filled,  while  the  head  should  have  a  diameter  of  from  1.6  to  2 
times  the  diameter  of  hole,  and  a  height  of  from  .6  to  .75  time 
the  hole  diameter.  Especial  care  should  be  taken  in  the  case 
of  machine-riveting  to  have  just  enough  metal  projecting  beyond 
the  hole  to  allow  for  the  necessary  upset  for  the  shank  to  fill 
the  hole,  with  just  enough  left  over  to  fill  the  die  for  the  head. 


B  C 

FIG.  63. 

TABLE  OF  DIMENSIONS  OF  RIVET  HEADS. 


Diameter 
of  Rivet. 

Pan  Head. 
A 

Button  Head. 
B 

Counter  Sunk. 
C 

d 

E 

F 

G 

E 

G 

E 

G 

f 

I 

i 

f 

I 

f 

If 

If 

| 

77.  Problem. — How  far  must  the  tail  of  the  rivet  project  in 
order  to  satisfy  the  above  conditions  for  the  following  case:  Two 
plates  f  inch  thick  each  are  to  be  connected,  using  f-inch  rivets 
in  H-inch  holes.  The  head  is  to  be 'cone-shaped,  having  an  out- 
side diameter  of  if  inches  and  a  height  of  f  inch. 

The  cubical  contents  of   the  cone  head  =  area    of 
altitude  =  2. 76  square  inches X. 25  inch  =  .69  cubic  inch. 


1  V 


- 


rflOl  SAJ  Q3T3VI8-  a^aUOO     *  - 


fcJi 


ml 


2.  ravjR  j-;; 


For  S1l.f!«.  Double.  Treble  Riveted  La?  Joint 
For  Single  Riveted  Double  Butt  Strap  Joint 

.    ••    Double 

.    ••    Treble     " 


&»- 


" 

.^^ 

^.'^.-' 

<x" 

^^^ 

1%-' 

^ 

„ 

^^ 

^ 

^$^ 

rr 

^r 

^ 

* 

x"" 

!    "7 

, 

r 

M 

«• 

',' 

X" 

*' 

*" 

*' 

, 

1" 

s 

i," 

i^l 

i  >«" 

Thickness  of  Plates-T. 
Fig.  1.    RIVET  DIAMETERS  BY  BACH'S  FORMULA. 


Fig.  3.    SINGLE  RIVETED  LAP  JOINT. 


Fig.  4.    DOUBLE  RIVETED  LAP  JOINT. 


Fig.  7.    DOUBLE  RIVETED  DOUBLE  BUTT  STRAP  JOINT..  Fig.  8.    DOUBLE  RIVETED  DOUB 


PLATE    I. 


II 


,     .  Thickness  of  Plates-T. 

Fig.  2.    RIVET  LENGTH  FOR  GIVEN  PLATE  THICKNESS 


fl.5.    DOUBLE  RIVETED  LAP  JOINT. 


Fig.  6.    TREBLE  RIVETED  LAP  JOINT. 


FT  STRAP  JOINT.  Fig.  9.    DOUBLE  RIVETiD  DOUBLE  BUTT  STRAP  JOINT. 


L 


.TVUOt  9AJ  031  3 


rSVto  SJ8UC 

\   1 


. 


RIVETED  JOINTS.  107 

The  difference  in  cubical  contents  between  a  hole  If  inch  in 
diameter  by  £  inch  long  and  a  shank  £  inch  in  diameter  and 
|  inch  long  =  £X.  7854(^2— |2)  =  .067  cubic  inch. 

The  amount  required  for  head  and  upset  therefore  equals 
.69 +  .067  =  .757  cubic  inch. 

The  area  of  the  f-inch  shank  =  .60  square  inch.  .757  cubic  inch, 

therefore,  calls  for  a  length  of  '—7-  =  1.25  inches.     This  amount 

would  then  be  the  projection  through  the  plate.     The  length  of 
rivet-shank  called  for  would  equal  f  inch  +  i|  inches  =  2  inches. 

NOTE. — Had  the  head  been  cup-shaped,  its  cubical  contents 
should  have  been  taken  as  that  of  a  spherical  segment.  For  cup- 
shaped  heads  the  diameter  is  about  i.  7  X  diameter  of  hole,  and 
the  height  about  .6  X  diameter  of  hole.  The  volume  of  the 
spherical  segment  is  given  by  the  following  rule:  Multiply  half 
the  height  of  the  segment  by  the  area  of  the  base  and  the  cube 
of  the  height  by  .5236  and  add  the  two  products. 

78.  Countersunk  Rivets. — Fig.  63^  shows  the  proportions  for 
a  countersunk  rivet.     Countersunk  rivets  make  a  much  weaker 
and  less  reliable  joint  than  the  ordinary  form,  and  should  only 
be  used  where  it  is  absolutely  necessary  that  the  surface  of  the 
plate  be  free  from  projections. 

79.  Nickel-steel  Rivets. — Where  peculiar  conditions   call  for 
great  strength  of  rivet  combined  with  small  area,  it  may  be  found 
desirable  to  use  nickel-steel  rivets.     Experiments  made  by  Mr. 
Maunsel  White  (see  Journal  Am.  Soc.  of  Nav.  Eng.,  1898)  on 
riveted  joints  using  nickel-steel  rivets  showed  an  average  shear- 
ing resistance  of  85,720  Ibs.  per  square  inch  for  single  shear,  and 
an  average  of  90,075  Ibs.  per  square  inch  for  double  shear.     These 
values,  it  will  be  noted,  are  nearly  double  those  of  the  very  mild 
steel  ordinarily  used.     The  rivets  were  f  inch  in  diameter,  and 
some  of  the  joints  failed  by  tearing  the  plates,  while  others  failed 
by  shearing  the  rivets. 


ic8  MACHINE  DESIGN. 

80.  Construction  of  Tight  Joints.  —  In  general  three  types 
of  riveted  joints  may  be  recognized : 

1.  Those  in  which  strength  is  the  sole  factor  of  importance, 
as  in  most  purely  structural  iron  and  steel  work. 

2.  Those  in  which  strength  and  tightness  are  equally  deter- 
mining elements,  as  in  boilers  and  pressure  pipes. 

3.  Those  in  which  tightness  is  the  prime  consideration,  as 
in  tanks  subjected  to  only  light  pressure. 

In  punchhg  holes  in  plates,  it  is,  of  course,  necessary  to 
have  th2  hole  in  the  die-block  larger  than  the  punch.  The 
consequence  is  that  the  holes  are  considerably  tapered  and 
care  should  be  exercised  in  joining  the  plate3  that  the 
small  ends  of  the  holes  be  together  as  shown  in  Fig.  64^, 
and  r.ot  apart  as  shown  in  Fig  64^.  It  is  obvious  that  at 
A  the  pressure  on  the  rivet  tends  to  draw  the  plates  closer 
tog.ther,  and  that  as  the  rivet  cools  its  longitudinal  shrinkage 
will  tend  to  keep  it  a  tight  fit  for  the  hole  in  spite  of  its  diametral 
shrinkage. 


FIG.  64.  FIG.  65. 

It  is  equally  obvious  that  at  B  the  pressure  on  the  rivet  will 
tend  to  force  the  plates  apart  and  squeeze  metal  between  them, 
and  also  that  all  shrinkage  of  the  rivet  will  be  away  from  the 
walls  of  the  hole. 

In  using  drilled  plates  care  must  be  exercised  to  remove  the 
sharp  burrs  left  by  the  drill,  as  experience  has  shown  that  this 
has  a  considerable  effect  on  the  strength  of  the  joint. 

Where  the  plates  form  the  walls  of  vessels  to  hold  fluids,  the 
joints  must  be  designed  with  a  view  toward  tightness  as  well  as 


RIVETED  JOINTS.  109 

strength.  For  this  purpose  the  edges  are  planed  at  a  slight  bevel, 
and  calked  as  shown  in  Fig.  65  by  a  tool  which  resembles  a  cold- 
chisel  with  a  round  nose.  Pneumatic  tools  are  used  for  this  pur- 
pose almost  entirely,  as  they  execute  more  uniform  and  rapid 
work  than  can  be  done  by  hand.  In  calking  great  care  should 
be  exercised  not  to  groove  the  plates  at  A- A,  as  these  are  danger- 
points  for  bending,  and  an  incipient  groove  is  very  apt  to  develop 
into  a  crack.  It  is  largely  on  this  account  that  the  round- 
nose  calking-tool  has  superseded  the  square-nose  in  the  best 
practice. 

It  has  been  found  that  the  load  which  the  joint  will  carry 
before  leaking  is  greatly  increased  if  both  rivet  heads  are  calked 
as  well  as  the  plate  edges.  (Bach's  experiments.) 

The  consideration  of  tightness  has  a  determining  effect  on  the 
maximum  allowable  pitch  for  any  given  thickness  of  plate  and 
type  of  joint.  Based  upon  practice  the  following  values  have 
been  found  safe  for  r5^  "  plates: — 

Single  riveted  lap  joints,  pitch  =  yt 
Double  riveted  lap  joints,  pitch  =  9.5t 
Double  riveted  butt  joints,  pitch  (in  outer  row)  =  14.5! 
Triple  riveted  butt  joints,  pitch  (in  outer  row)  =  2ot. 
Because  of  the  use,  with  heavier  plates,  of  rivet  diameterv 
which  are  proportionately  too  small,   these  ratios  of  pitch  t«» 
thickness  of  plate  will  be  found  to  decrease  in  practice  as  the 
thickness  of  plate  increases.     Thus  for  \"  plates  they  become 
5t,  6.6t,  10.251  and  i6t,  respectively. 

8 1.  Materials  to  be  Used. — The  material  to  be  used  in  riveted 
joints  depends,  of  course,  on  the  nature  of  the  work,  but  in  gen- 
eral it  may  be  said  that  extremely  mild  and  highly  ductile  steel 
as  free  from  phosphorus  and  sulphur  as  possible  should  be  used. 
Open -hearth  steel  is  greatly  to  be  preferred  to  Bessemer.* 

*  Standard  specifications  can  be  found  in  Kent's  ''Mechanical  Eng.  Pocket 
Book."  The  U  S.  Navy  Dept.  has  adopted  standard  rules  for  riveting  naval 
vessels.  These  can  be  found  in  Marine  Engineering,  Jan.,  1898. 


I  io  MACHINE  DESIGN. 

82.  Plates  with  Upset  Edges.—  Some  boiler-makers  have 
adopted,  as  an  expedient  for  saving  material,  a  method  of  using 
plates  with  thickened  (upset)  edges.  If  we  let  t  represent  the 
thickness  of  the  body  of  the  plate  and  t'  the  thickness  of  the  edge, 
while  a  represents  the  pitch  and  d  the  diameter  of  hole,  then,  when 


the  joint  will  be  as  strong  as  any  other  section  of  the  plate,  the 
joint  being  proportioned,  of  course,  for  the  thickness  t'.  It  is 
customary  to  thicken  only  the  edges  which  form  the  longitudinal 
seam.  This  method  is  open  to  two  serious  objections.  Unless 
the  plates  are  very  carefully  annealed  after  being  upset  they  are 
almost  certain  to  be  weakened  by  indeterminate  working  and 
cooling  stresses.  Moreover,  although  the  original  new  joint  may 
show  as  high  an  efficiency  as  if  the  plates  throughout  were  of  the 
thickness  I',  as  corrosion  proceeds,  it  acts  more  on  the  plate  away 
from  the  joint  than  at  the  joint,  because  at  the  latter  place  the 
plate  is  protected  by  the  cover-plate  or  rivet-head  or  both.  The 
result  is  a  shorter  life  under  full  pressure  for  the  boiler  with  thin 
plates  and  thickened  edges. 

83.  Joints  for  More  than  Two  Plates.  —  The  joints  considered 
thus  far  have  dealt  with  the  problem  of  connecting  the  edges 
of  two  plates  only.  In  tanks  and  boilers  which  must  have  tight 
seams  we  are  frequently  confronted  with  the  problem  of  joining 
three  and  even  four  plates.  An  instance  is  where  the  cross-seam 
and  longitudinal  seam  of  a  boiler  meet.  The  joint  is  made  by 
thinning  down  one  or  more  of  the  plates.  Figs.  66  to  70  (taken 
from  Unwin's  "Machine  Design")  show  the  methods  employed  . 
Fig.  66  shows  a  junction  of  three  plates,  a,  b,  and  c,  where  both 
seams  are  single-riveted  lap-joints.  It  will  be  seen  that  the 
corner  of  a  is  simply  drawn  down  to  an  edge  and  "tucked 
under"  c. 

Fig.  67  shows  a  junction  of  three  plates  where  one  seam  is  a 


RIVETED  JOINTS. 


single-riveted  and    the    other   a    double-riveted    Jap-joint.      As 
before,  the  corner  of  a  is  drawn  down  and  tucked  under  c. 

Fig.  68  shows  the  junction  of  three  plates  where  both  seams 
are  single -riveted,  single  cover-plate  butt-joints.  The  plates 
merely  abut  against  each  other,  but  the  longitudinal  cover  is 


FIG.  66. 


FIG.  68. 


drawn  down  and  tucked  under  the  cross-seam  cover  which 
thinned  down  to  match. 


FIG.  69.  FIG.  70. 

Fig.  69  also  shows  the  junction  of  three  plates ;  here  the  cross- 
seam  is  a  single-riveted  lap-joint  while  the  longitudinal  joint 
is  a  double-riveted  butt-joint  with  double  cover-plates.  The 
upper  cover-plate  is  planed  on  the  end  so  that  it  can  be  tightly 
calked  where  it  abuts  against  the  plate  c. 

A  method  of  joining  four  plates  is  shown  in  Fig.  70.  Both 
seams  are  single-riveted  lap-joints,  b  and  c  are  both  drawn 
down  as  shown. 

84.  Junction  of  Plates  Not  in  Same  Plane. — Where  the  plates 
to  be  joined  are  in  different  planes,  it  is  customary  to  use  some 
one  of  the  rolled  structural  forms.  Fig.  71  shows  the  method 
of  using  an  angle  iron  for  plates  at  a  right  angle  to  each  other. 


MACHINE  DESIGN. 


Where  it  is  possible  to  turn  a  flange  on  one  of  the  plates  this 
method  is  often  adopted.     Care  should  be  taken  not  to  use  too 


FIG.  71. 


sharp  a  radius  of  curvature  (the  inside  radius  must  be  greater 
than  the  thickness  of  the  plate  even  with  the  mildest  steel)  and 
the  flanged  plate  should  be  thoroughly  annealed  after  it  is  bent. 

Fig.  72  shows  the  method  of  making  flanged  joints  such  as 
are  frequently  used  in  connecting  boiler  heads  and  shells. 

85.  Problem. — The  following  problem  will  serve  to  illustrate 
the  design  of  riveted  joints  for  boilers.  It  is  required  to  design  a 
horizontal  tubular  boiler  48  inches  in  diameter  to  carry  a  work- 
ing pressure  of  100  pounds  per  square  inch. 

A  boiler  of  this  type  consists  of  a  cylindrical  shell  of  wrought 
iron  or  steel  plates  made  up  in  length  of  two  or  more  courses  or 
sections.  Each  course  is  made  by  rolling  a  flat  sheet  into  a 
hollow  cylinder  and  joining  its  edges  by  means  of  a  riveted  joint, 
called  the  longitudinal  joint  or  seam.  The  courses  are  joined  to 
each  other  also  by  riveted  joints,  called  circular  joints  or  cross- 
seams.  Circular  heads  of  the  same  material  have  a  flange  turned 
all  around  their  circumference,  by  means  of  which  they  are 
riveted  to  the  shell.  The  proper  thickness  of  plate  may  be 
determined  from  (I)  The  diameter  of  shell  =  48  inches;  (II)  The 
working  steam-pressure  per  square  inch  =  ioo  pounds;  (III)  The 
tensile  strength  of  the  material  used;  let  steel  plates  be  used 
of  60,000  pounds  specified  tensile  strength. 

Preliminary  investigations  Qf  he  conditions  of  stress  in  the 
cross-section  of  material  cut  by  u,  plane  (I)  Through  the  axis; 
(II)  At  right  angles  to  the  axis,  of  a  thin  hollow  cylinder,  the 
stress  being  due  to  the  excess  of  internal  pressure  per  square  inch. 


RIVETED  JOINTS.  113 

Let  Z,  =  the  length  of  the  cylindrical  shell  in  inches; 
D  =  ihe  diameter  of  the  cylindrical  shell  in  inches; 
^>  =  the  excess  of  internal  over  external  pressure  in  pounds 

per  square  inch ; 
/i  =  unit  tensile  stress  in  a    longitudinal  section  of  material 

of  the  shell  due  to  P; 
/2  =  unit  tensile  stress  in  a  circular  section  of  material  of  the 

shell  due  to  p; 
t  =  thickness  of  plate; 
ft  =  ultimate  tensile  strength  of  plate. 
All  stresses  are  in  pounds  per  square  inch. 

In  a  longitudinal  section  the  total  stress  is  equal  to  LDp, 

Dp 

and  the  area  of  metal  sustaining  it  =  2/J.     Then  /i=~^7- 

In  a  circular  section  the  total  stress  = ,   and  the  area 

4 

sustaining  \t=xDt,  nearly.     Then 

xD2p       i       Dp 

t —  Y — 

/2~    4     *xDt      #' 

Therefore  the  stress  in  the  first  case  is  twice  as  great  as  in 
the  second;  and  a  thin  hollow  cylinder  is  twice  as  strong  to 
resist  rupture  on  a  circular  section  as  on  a  longitudinal  one. 
The  latter  only,  therefore,  need  be  considered  in  determining  the 
thickness  of  plate.  Equating  the  stress  due  to  p  in  a  longitudinal 
section,  and  the  strength  of  the  cross-section  of  plate  that  sustains 

it,  we  have  LDp  =  2Ltft.    Therefore  /= — T-,  the  thickness  of  plate 

that  would  just  yield  to  the  unit  pressure  p.  To  get  safe  thickness, 
a  factor  of  safety  K  must  be  used.  It  is  usually  equal  in  boiler- 
shells  to  5  or  6.  Its  value  is  small  because  the  material  is  highly 
resilient  and  the  changes  of  pressure  are  gradual,  i.e.,  there  are 
no  shocks.  This  takes  no  account  of  the  riveted  joint,  which 


H4  MACHINE  DESIGN. 

is  the  weakest  longitudinal  section,  E  times  as  strong  as  the 
solid  plate,  E  being  the  joint  efficiency  =0.75   if  the  joint  be 

double-riveted.       The    formula    then    becomes    t=    '  £.    Sub- 
stituting values, 

6X48X100 

t  = , =0.32  inch,  say  A  inch. 

2X60000X0.75 

The  circular  joints  will  be  single-riveted  and  joint  efficiency 
will  =0.50.  But  the  stress  is  only  one  half  as  great  as  in  the 
longitudinal  joint,  and  therefore  it  is  stronger  in  the  proportion 
0.50X2  to  0.75,  or  i  to  0.75.  From  this  it  is  seen  that  a  circular 
joint  whose  efficiency  is  0.50  is  as  strong  as  the  solid  plate  in  a 
longitudinal  section.  From  the  value  of  /  the  joints  may  now  be 
designed. 

Consider  first  the  cross-seam.  This  is  a  single-riveted  lap- 
joint.  Assume  drilled  holes. 

Equations  (5), 

,=,4, 


and  (6),  , 

apply,  while  from  Table  II  we  get  as  values  of  ft,  /«,  fc,  60,000, 
45,000,  and  85,000  pounds  per  square  inch  respectively  for  steel 
plates  and  rivets. 

85000 


45000 


=.75  inch 


•.5  X45000 
and         a  =       -3I25X6oooo      +  -75  =  1-81  mches,  say  itf  inches. 

The  margin  =d  =  .75  inch. 

The  lap  of  the  cross-seam  =3^  =  2.  2  5  inches. 

The  longitudinal  seam  will  be  a  double  staggered  lap-joint. 


RIVE  TED  JOINTS.  !  x  5 

Equations  (n)  and  (12)  apply: 


.2  and    a- 

h 


From  Table  II,  ft  =60,000,  /,  =46,000,  and  fc  =85,000; 

85000 
—  say  .75, 


i. 57 X. 75  X46ooo 
and       a  =     >3I25x6oooo     +  -75  =  2-93  inches,  say  2«  inches. 

The  distance  between  the  rows  =  1.88^  =  1.41  inches,  say  i& 
inches.  The  total  lap  in  the  longitudinal  joint  =4.88^  =  3.66 
inches,  say  3}$  inches. 

The  joints  are  therefore  completely  determined,  and  a  detail 
of  each,  giving  dimensions,  may  be  drawn  for  the  use  of  the  work- 
men who  make  the  templets  and  lay  out  the  sheets. 

Having  determined  the  proportion  of  the  joints,  let  these 
dimensions  be  used  to  calculate  the  actual  efficiency  of  the  longi- 
tudinal seam. 

Assume  that  the  natural  tensile  strength  of  the  unperforated 
plate  is  60,000  Ibs.  per  square  inch. 

The  excess  of  strength  of  drilled  steel  plates  in  net  section 
over  unperforated  section  (see  §  63)  is 

/      6.i24\/4.5-r\ 

2  + — 7^  )   — - —  Jper  cent. 
\        Vt  A    2.5  /F 

Here        £  =  .3125  in.    and    r=^-=2'9^      =3.92; 
.'.  excess  due  to  perforation  =3%,  nearly. 


MACHINE  DESIGN. 

6o,000  X  1  .03  =  6  1  ,800  =  ft. 
f,  =46,000  from  Table  II. 

/c=85,ooo    »          »      " 

r  =  (a-d)//,  =  (2.9375-.75)(.3i25X6i,8oo)  =42,250  Ibs. 
S  =  i.57rf2/s==i.57X.752X46,ooo=4o,625lbs. 


55,075  Ibs. 
Of  T,  5,  and  C,  the  latter  has  the  smallest  value;  the  actual 

efficiency  of  the  joint  may  be  taken  as  =  p  =  —    —  =  -7235    or 

72-35%. 

Since  T1,  S,  and  C  are  unequal  it  is  evident  that  there  has  been 
departure  from  the  conditions  for  maximum  efficiency.  There  are 
two  ways  of  restoring  this  equality,  or  at  least  diminishing  the 
inequality.  If  a  be  slightly  decreased,  Tand  P  will  be  propor- 
tionately decreased  and  S  and  C  will  have  the  same  values  as 
before. 

Leaving  a  as  before  and  increasing  d,  increases  5  as  the 
square  of  d  and  C  as  d,  while  T  and  P  remain  as  before. 

Inspection  shows  that  T  exceeds  C  by  5.7  per  cent.  There- 
fore a  may  be  decreased  by  this  percentage,  or  .17  inch.  This 
is  approximately  &  inch  and  reduces  the  pitch  from  att  to 
2f  inches. 

Using  this  value  of  a  gives  as  the  excess  strength  due  to  per- 

foration 4.33  per  cent. 

/.  ft  =  62,600  Ibs., 

T  =  39,i25  Ibs., 
5  =  40,625  Ibs., 
C  =  39,845  Ibs., 
P  =  Si,S6o  Ibs., 
T  39125 


The  second  method  of  balancing  T,  S,  and  C  would  be  by 


RIVETED  JOINTS.  1  1  ^ 

increasing  d.  The  next  commercial  size  above  f  inch  would 
be  H  inch.  Leave  a  =  2$  inches,  and  increase  d  to  H  inch,  and 
first  calculate  the  excess  strength  due  to  perforation. 


Using  these  values,  the  excess  =4.57  per  cent. 

^=62,750  Ibs., 
T  =  41,670  Ibs., 
5  =  47,675  lbs-, 
C  =  43,i65  lbs., 
P  =  55,075  lbs., 
and,  since  T  is  least, 

T     41*70         6 
P     55075 

The  best  result  is  that  obtained  by  keeping  d  =  f  inch,  but 
changing  a  to  2f  inches,  and  it  would  be  advantageous  to  make 
these  the  proportions  of  the  joints  rather  than  those  first  de- 
termined. 

The  next  step  is  to  check  back,  using  the  proportions  de- 
cided upon,  for  the  actual  factor  of  safety  which  should  not  be 
less  than  5. 

T7"  r\  L 

From  the  equation  (p.  107)  t  =  ~r^  ,  we  have 


.  .- 

48X100 

Attention  should  be  called  to  the  fact  that  the  ordinary  and 
less  correct  method  of  calculating  efficiencies  ignores  the  excess 


Ii8  MACHINE  DESIGN. 

strength  due  to  perforation,,  and  the  efficiency  is  simply  taken 
T_ 

With  a=2if  inches,  and  J  =  |  inch,  this  would  give  us 
77    4I°15  =  74-47%>  instead  of  72.35%. 


With  o  =  af  inches  and  </  =  £  inch,  it  would1  give  us 
72'7I%'  instead  °f  75-88%. 


With  a=2&  inches  and  d=$  inch,  it  would  give 
=  7i.97%>  instead  of  75.66%. 


CHAPTER  VIII. 

BOLTS     AND     SCREWS. 
86.  Classification    and  Definition. — Bolts  .and  screws  may  be 

^fU^ 

classified  as  follows:  I.  Bolts;  II.  Stud$;  III.  Cap-screws,  or 
Tap-bolts;  IV.  Set-screws;  V.  Machine  screws;  VI.  Screws 
for  power  transmission. 

A  "bolt"  consists  of  a  head  and  round  body  on  which  a 
thread  is  cut,  and  upon  which  a  nut  is  screwed.  When  a  bolt 
is  used  to  connect  machine  parts,  a  hole  the  size  of  the  body  of 
the  bolt  is  drilled  entirely  through  both  parts,  the  bolt  is  put 
through,  and  the  nut  screwed  down  upon  the  washer.  (See 
Fig.  73-) 


FIG.  73. 


FIG.  74. 


FIG.  75. 


A  "stud"  is  a  piece  of  round  metal  with  a  thread  cut  upon 
each  end.  One  end  is  screwed  into  a  tapped  hole  in  some  part 
of  a  machine,  and  the  piece  to  be  held  against  it,  having  a  hole 
the  size  of  the  body  of  the  stud,  is  put  on  and  a  nut  is  screwed 
upon  the  other  end  of  the  stud  against  the  piece  to  be  held.  (See 

?ig.  74-) 

119 


izo  MACHINE  DESIGN. 

A  "cap-screw"  is  a  substitute  for  a  stud,  and  consists  of  a 
head  and  body  on  which  a  thread  is  cut.  (See  Fig.  75.)  The 
screw  is  passed  through  the  removable  part  and  screwed  into  a 
tapped  hole  in  the  part  to  which  it  is  attached.  A  cap-screw 
is  a  stud  with  a  head  substituted  for  the  nut. 

A  hole  should  never  be  tapped  into  a  cast-iron  machine 
part  when  it  can  be  avoided.  Cast  iron  is  not  good  material 
for  the  thread  of  a  nut,  since  it  is  weak  and  brittle  and  tends  to 
crumble.  In  very  many  cases,  however,  it  is  absolutely  neces- 
sary to  tap  into  cast  iron.  It  is  then  better  to  use  studs  if  the 
attached  part  needs  to  be  removed  often,  because  studs  are  put 
in  once  for  all,  and  the  cast-iron  thread  would  be  worn  out 
eventually  if  cap-screws  were  used. 

The  form  of  the  United  States  standard  screw-thread  is 
shown  in  Fig.  76.  The  sides  of  the  thread  make  an  angle  of 


FIG.  76. 

60°.  Instead  of  coming  to  a  sharp  point,  the  threads  have  a 
flat  at  top  and  bottom  whose  width  is  =  |/>,  p  being  the  pitch. 

Table  VI  gives  the  standard  proportions. 

When  one  machine  part  surrounds  another,  as  a  pulley -hub 
surrounds  a  shaft,  relative  motion  of  the  two  is  often  prevented 
by  means  of  a  "set-screw,"  which  is  a  threaded  body  with 
a  small  square  head  (Fig.  77).  The  end  is  either  rounded  as 
in  Fig.  77  a,  or  pointed  as  in  Fig.  77  6,  or  cupped  as  in  Fig. 
77  c,  and  is  forced  against  the  inner  part  by  screwing  through 
a  tapped  hole  in  the  outer  part. 

Data  relative  to  the  holding  power  of  set-screws  will  be 
found  in  §  109. 


BOLTS  AND  SCREWS. 


TABLE  VI. — U.  S.  STANDARD  SCREW-THREADS. 


Bolts  and  Threads. 

Hex.  Nuts  and  Heads. 

AT  and'  H. 

i 

1 

• 

"3 

•g 

tf 

1 

1 

I 

1 

1 

! 

b 

"o-g 

1 

« 

I. 

1 

1 

1 

pj 

1 

1 

fcg 

IM 

Q   ' 

3  . 

Q  • 

i 

8 

Q  * 

I 

H 

•gi 

Jo 

Width  c 

jfl 

1 

j! 

i 

!«! 

'I 

a 

I 

j 

1*. 

Ins. 

Ins. 

Sq.  Ins.'Sq.  Ins. 

Ins. 

Ins. 

Ins. 

Ins. 

Ins. 

Ins. 

i 

20 

18 

•185 

.240 

.0062 

.0074 

.049 
•077 

.027 
-045 

tl 

f 

I 

4 

A 

ft 

t 

16 

.294 

.0078 

.110 

.068 

H 

A 

ff 

i 

14 

•344 

.0089 

.150 

•093 

fl 

li 

TV 

A 

I 

Z?V 

i 

13 

.400 

.0096 

.196 

.126 

1 

ll 

I 

i 

A 

41 

\ 

12 

•454 

.0104 

.249 

.162 

B 

M 

l| 

A 

i 

i  ii 

II 

•5°7 

.0113 

•307 

.  202 

*A 

I^j 

f 

A 

i| 

IO 

.620 

.0125 

.442 

.302 

it 

JA 

f 

tt 

iti" 

r 

I 

:& 

.0138 
.0156 

.601 

.785 

.420 
•550 

J 

3r 

If 
JH 

3 

7 

.940 

.0178 

•994 

.694 

IT! 

j 

2^y 

i 

A 

2A 

7 

.065 

.0178 

.227 

•893 

2 

if 

2A 

i 

A 

2  |f 

6 

.0208 

•485 

•057 

2A 

| 

2M 

i 

A 

3^j 

6 

^284 

.0208 

.767 

•295 

2i 

2A  '  2f 

i 

A 

3  1| 

si 

-389 

.0227 

-074 

•515 

2A 

2! 

2§i 

1 

A 

3f 

| 

5 

.491 

.0250 

-405 

-746 

2f 

2ri 

3A 

if 

tt 

sU 

[ 

t 

41 

4 

.616 

.712 
.962 
.176 

.0250 
.0277 
.0277 
.0312 

.761 
3-I42 
3-976 
4-9°9 

.051 
.302 
3-023 
3-7I9 

! 

1 

1 

il 

1 

al 
i 

I 

4 

.426 

.0312 

5-940 

4.620 

4l 

4A 

4» 

2f 

2H 

6 

3i        -629 

•0357 

7.069 

5.428 

4f 

4A      5f 

3 

2rt 

6$3 

34 

•  879 

•0357 

8.296 

6.  510 

5 

3A 

7tV 

3i 

3.100 

.0384 

9.621 

7-548 

5t 

sA 

6^-      3 

3A 

7ff 

3 

3-317 

.0413 

11.045 

8.641 

Si 

stt 

6§i 

3 

3tt 

H 

3 

3-567 

.0413 

12.566 

9-963 

6i 

6A 

7A 

4 

3tt 

8ft 

\ 

3-798 

•0435 

14.186 

11.329 

69 

6A 

7A 

4i 

4A 

9A 

2i 

4.028 

•0454 

15.904 

12-753 

6f 

6rf 

7§i 

4i 

4A 

9f 

\ 

2i 
2i 

4.256 
4.480 

.0476 
.0500 

17.721 
I9-635 

14.226 
15  •  763 

?1 

7A 

i 

5 

4H 

10" 

k 

2* 

4-730 

.0500 

21.648 

I7-572 

8 

7lf 

9* 

Si 

sA 

"H 

1 

2I 
2I 

4-953 

5-2o3 

.0526 
.0526 

25.967 

19.267 
21.262 

8|        8A    ^§f 

9 

stt    i"f 

al 

•0555 

28.274 

23.098 

9A 

1035 

6 

5«   !    "» 

MACHINE  DESIGN. 


The  term  "machine  screws"  covers  many  forms  of  small 

screws,  usually  with  screw-driver  heads.  All  of  the  kinds  given 

in  this  classification  are  made  in  great  variety  of  size,   form, 
length^  etc. 

miim^^ 


FIG.  77.  FIG.  78. 

Thus  far  American  manufacturers  have  failed  to  agree  upon 
standard  dimensions  for  set-screws  and  machine  screws.* 

For  consideration  of  their  design,  we  will  divide  bolts  and 
screws  into  three  classes : 

(a)  Those  which  are  put  under  no  stress  by  screwing  up. 

(b)  Those  which  are  put  under  an  initial  stress  by  tightening. 

(c)  Those  which  are  used  to  transmit  power. 

87.  Analysis  of  Action  of  Screw. — Before  taking  these  cases 
up  in  detail  ii  will  be  well  to  examine  into  the  general  action 
of  screw  and  nut.  Reference  is  made  to  Fig.  79.  The  turning 
of  a  nut  loaded  with  W  Ibs.  may  be  considered  as  equivalent 
to  moving  a  load  W  on  an  inclined  plane  whose  angle  with 
the  horizontal  is  the  same  as  the  mean  pitch  angle  of  the 
thread  a. 

*  The  report  of  the  committee  of  the  A.  S.  M.  E.  on  this  subject  with  suggested 
standards,  will  be  found  in  Vol.  28  of  the  Trans.  A.  S.  M.  E. 


BOLTS  AND  SCREWS.  123 

Let  r\=  outside  radius  of  thread; 
r2  =  inside  radius  of  thread; 

r  =  mean  radius  of  thread,  approximately  —    — ; 

p  =  pitch  of  thread; 

a  =  mean  pitch  angle,  i.e.,  tan  a  =—  ; 

2TCT 

fj.  =  coefficient  of  friction  between  nut  and  thread; 

(p  =  angle  of  friction  between  nut  and  thread,  i.e.,  tan  <f>  =  ft. 

ist.  To  raise  W. 

Consider  W  as  a  free  body  moving  uniformly  up  the  incline 
under  the  system  of  forces  shown  in  Fig.  80,  where  W  =  axial 


FIG.  79.  FIG.  80. 

load,  R  the  normal  reaction  between  nut  and  thread  due  to  W, 
H  the  horizontal  push  forcing  the  nut  up  the  incline,  and  F 
the  friction  in  direction  of  incline  due  to  the  normal  pressure  R; 
then,  by  the  ordinary  laws  of  mechanics, 

R  =  W  +  cosa, (i) 

F  =  fjLW  +  cos  a, (2) 

H  =  Wtzn  («  +  </>) (3) 

The  turning  moment  M 

=  Hr  =  Wr  tan  («  +  <£) (4) 


124 


MACHINE  DESIGN. 


Since  tana  =  —  ,  and  tan  <f>  =  fi,  while 


,. 
tan(a  +  <£) 


tan  a  4-  tan  <j> 
-  -  -  -  —  ^, 
i—  tan  a  tan  < 


it  follows  that 


2d.  To  lower  PF, 


(c) 

(6) 
(  } 


The  foregoing  has  applied  to  square  threads.  Consider  V 
threads  with  /9  =  the  angle  of  V  with  a  plane  normal  to  the  axis 
of  the  screw.  (See  Fig.  81.)  The  mean  helix  angle  =  a  as  be- 


FIG.  81. 


fore,  but  now  R  slopes  from  W  in  two  directions,  making  the 
angle  a  in  the  one,  as  before,  but  also  making  the  angle  /?  with 
W  in  a  plane  at  right  angles  with  the  first.  Hence 


^/jiW  sec  a  sec/? 


(8) 
(9) 


For  raising  load, 


BOLTS  AND  SCREWS. 


125 


For  lowering  load, 


Hr=Wr1- 


sec 


88.  Calculation  for  Screws  Not  Stressed  in  Screwing  Up. — 
Returning  now  to  (a).  As  illustrations  of  this  class  consider 
the  eye-bolts  shown  in  Fig.  82.  It  is  customary  to  neglect  the 
influence  of  the  thread  on  the  strength  of  the  bolt,  and  to  con- 
sider as  the  effective  area,  A,  to  resist  stress,  only  the  area  of  a 
circle  whose  diameter  equals  the  diameter  of  the  bolt  at  the 
bottom  of  the  thread.  In  both  cases  considered  a  torsional 
stress  is  induced  by  screwing  the  engaging  surfaces  together, 
but  if  these  surfaces  are  a  proper  fit,  this  stress  is  negligible, 
particularly  since  it  exists  only  within  the  limits  of  the  engaging 
threads  where  the  action  of  the  further  working  load  which  the 
bolt  bears  does  not  come  into  play. 


FIG.  82. 


I 

T 

FIG.  83. 


The  eye-bolt  being  now  subjected  to  the  working  load  T  in 
the  direction  of  its  axis,  we  have 


where  /  is  the  safe  unit  working  stress  for  the  material  and  con- 
ditions. 

If  U  =  ultimate  unit  strength  of  the  material,  then,  if  the 


126  MACHINE  DESIGN. 

load  is  a  constant,  dead  load,  /  may  be  taken  as  great  as  —  for 
good  wrought  iron  or  mild  steel.  If  the  load  is  a  variable  one, 
slowly  applied  and  removed,  /  should  not  exceed  —  for  the 
same  materials.  If  the  load  is  variable  and  suddenly  applied, 
/  should  never  exceed  —  for  these  materials,  and  in  cases  of 

shock  may  need  to  be  much  smaller  than  this. 

The  case  shown  in  Fig.  83  must  not  be  confused  with  the 
preceding.  Here  (as  will  be  explained  under  (b) )  there  may  be 
a  tensile  stress  in  A  induced  by  compressing  B-B  between  the 
shoulder  and  the  nut.  If  the  extension  in  the  part  A  of  the 
bolt  due  to  the  application  of  the  force  T  later  be  greater  than 
the  compression  caused  in  B-B  by  the  tightening  of  the  nut, 
then  the  shoulder  will  leave  B-B,  and  simple  tension  =  T  results 
in  all  sections  of  the  bolt  below  the  nut,  as  in  case  (a). 

On  the  other  hand,  if  the  extension  in  the  part  A  of  the 
bolt  due  to  the  subsequent  application  of  T  be  not  so  great  as 
the  original  compression  of  B-B  due  to  the  screwing  up,  then 
we  have  in  the  part  A  a  resultant  tension  greater  than  T.  This 
case  would  come  under  (b}. 

89.  Calculation  of  Screws  Stressed  in  Screwing  Up. — (b) 
Combined  tension  and  torsion  are  induced  in  a  bolt  by  tighten- 
ing it.  The  stress  may  equal,  or  very  greatly  exceed,  the  tensile 
stress  due  to  working  forces.  Consider  the  example  shown  in 
Fig.  78.  Suppose  the  nut  screwed  down  so  that  the  parts  con- 
nected by  the  bolt  are  held  close  together  at  E-F  but  have  not 
yet  been  compressed.  Suppose  that  the  proportions  are  such 
that  the  wrench  may  be  given  another  complete  turn.  The  nut 
will  move  along  the  direction  of  axis  of  the  bolt  a  distance  =  p. 
The  parts  held  between  the  head  and  nut  will  be  compressed 
and  the  body  of  the  bolt  will  be  extended. 


DOLTS  AND  SCREWS.  127 

The  force  applied  at  the  point  B,  or  end  of  the  wrench 
(a  distance  /  from  the  axis)  will  range  from  a  value  of  o  at  the 
beginning  of  the  turn  to  a  value  P  at  the  finish.  The  average 

p 
value  of  the  turning  force  will  be  approximately  =  — . 

This  distance  moved  through  by  the  point  of  application  of 
the  force  is  2~L  Hence  the  work  done  in  turning  the  nut  a 
full  turn  under  these  conditions  will  be 


-.2rd  =  Pxl (12) 


The  resistances  overcome  by  this  application  of  energy  are 
three  in  number: 

ist.  The  work  done  in  extending  the  bolt. 

2d.  The  work  done  in  overcoming  the  frictional  resistance 
between  nut  and  thread. 

3d.  The  work  done  in  overcoming  the  frictional  resistance 
between  nut  and  washer. 

These  will  be  considered  in  order. 

ist.  Let  T'  =  the -final  pure  tensile  stress  in  the  bolt  due  to 
screwing  up  one  turn.  At  the  beginning  of  the  turn  the  ten- 

T 

sion=o.     The  average  value  may  be  considered  =—  for  the 

turn.  The  distance  moved  through  by  the  point  of  application 
of  this  force  in  the  direction  of  its  line  of  action,  in  one  turn  =  ^. 
The  work  done  in  extending  the  bolt 

-7* (13) 

2d.  The  frictional  resistance  between  the  threads  of  nut  and 
bolt  depends  upon  the  form  of  the  thread  as  well  as  the  mate- 
rials used  and  the  condition  of  the  surfaces.  (See  equations  (2) 

^    ^*A— > 

*" 


128  MACHINE  DESIGN. 

and  (9),  §  87.)     Assuming  a  V  thread  as  being  more  commonly 
used  for  fastenings,  the  average  value  of  the  friction 

T 

F  =  fi —  sec  a  sec  ft 


T 

(eq.  (9) ),  since  the  average  load  for  the  turn  =  — . 

The  distance  moved  through  by  the  point  of  application  of 
F  for  one  turn  of  the  nut  on  the  bolt  =  ^>  cosec  a.  Hence  the 
work  done  in  overcoming  the  friction  between  bolt  and  nut  in 

the  one  turn  =  /<—  sec  a  sec  ft  .  p  cosec  a 

= —  ftp  sec  a  sec  ft  .  cosec  a (14) 


3d.  The  frictional  resistance  between  nut  and  washer  due  to 

T  T 

a  mean  force  —  will  be  pf — ,  in  which  //  is  the  coefficient  of 

friction  between  nut  and  washer.  The  point  of  application  of 
this  resistance  may  be  taken  at  a  distance  of  -fi  from  the  axis 

of  the  bolt,  TI  being  the  outside  radius  of  bolt-thread.  The  dis- 
tance moved  through  by  the  point  of  application  for  one  turn  of 
the  nut  =  27rfri,  and  the  work  done  in  overcoming  this  frictional 
resistance 

=-//>n (15) 

Equating  (12)  to  the  sum  of  (13),  (14),  and  (15),  gives 


T       T  T 

Pnl=—p+—[ip  sec  a  sec  ft  cosec  a-^—pf 


BOLTS  AND  SCREWS. 


129 


whence 


p+fipsec  a  sec  /?  cosec  a 


'      '     *     " 


/*=-T  =  unit  stress  in  bolt  due  to  pure  tension.  .     (17) 

In  addition  to  this  it  must  be  borne  in  mind  that  the  sere™ 
is  subjected  to  a  torsional  moment  whose  value  can  be  deter 
mined  by  considering  the  nut  as  a  free  body  as, 
shown  in  plan  view  in  Fig.  84,  where  all  of  the 
forces  capable  of  producing  moments  about  the 
axis  of  the  nut  are  indicated  as  they  exist  at  the 
end  of  the  turn. 

Summing  the  moments  about  the  axis  of  the 
bolt  gives 

FIG.  84.  Hr-Pl-t/Tfr  .....     (18) 

Hr  is,  of  course,  the  torsional  moment  transmitted  from  the  nut 
to  the  bolt.  To  find  its  numerical  value  substitute  the  value  of 
T  found  in  equation  (16)  and  solve  (18). 

The  unit  stress  induced  in  the  outer  fibers  of  a  rod  of  cir- 
cular section  and  radius  r2  (  =  radius  at  bottom  of  thread)  is 
found  by  means  of  the  equation 

}>-  =  M  ........     .     (19) 

J  is  the  polar  moment  of  inertia,  in  this  case  =  —  —  ;  c  is  the  dis- 

tance from  neutral  axis  to  most  strained  fiber,  in  this  case  =  fa  > 
f8  is  the  induced  unit  stress  in  outer  fiber;  M  is  the  moment,  in 
this  case  =  #>.  Combining  equations  (18)  and  (19)  and  sub- 
stituting these  values  gives 


/•= 


(20) 


MACHINE  DESIGN. 


The  equivalent  tensile  unit  stress  due  to  the  combined  action 
of  }t  and  fa  is  found  from  the  equation  for  combined  tension 
and  torsion, 

(21) 


go.  Problem.  —  What  is  the  unit  fiber  stress  induced  in  a 
U.  S.  standard  J-inch  bolt  in  screwing  up  the  nut  with  a  pull  of 
one  pound  at  the  end  of  a  wrench  8  inches  long  ?  Arrangement 
of  parts  as  shown  in  Fig.  78. 

In  this  case  d\  =  .  500  in.,        ri  =  .25  in., 

d2  =  .400  in.,        r2  =  .2  in., 
r=.225  in.,        A  =.126  sq.  in., 
^=.077  in., 
^  =  ^'  =  0.15, 

P 

a  =  angle  whose  tangent  is  —  =  3°  7', 

sec  a  =  1.0015,          cosec  a  =  18.39, 
/?  =  3o°,  sec  /?=  1.155, 

P  =  ilb.,        and          J  =  8ins. 

From  equation  (16) 

__  2XIX7TX8  _  _____ 

".077  +  .077X0.15X1.0015X1.  155X18.39+0.15X3X^X0.25 
=  74.467  Ibs. 
From  equation  (17), 


From  equation  (20), 

.  =  2(1X8-0.15X74-467x1X0.25) 

=  213  Ibs. 


BOLTS  AND  SCREWS.  13 i 

From  equation  (21), 


=  691.4  Ibs. 

If  a  pull  of  one  pound  on  an  8-inch  wrench  applied  to  a  \- 
inch  bolt  can  induce  a  unit  fiber  stress  of  691.4  Ibs.,  since  equa- 
tions (16)  and  (20)  show  that  the  stress  increases  directly  as  the 
pull,  it  follows  that  a  pull  of  30  Ibs.,  such  as  is  readily  exerted 
by  a  workman,  will  induce  a  stress  of  30X691.4  =  20,742  Ibs. 
per  square  inch. 

91.  Wrench  Pull. — If  this  turning  up  be  gradual  and  the 
bolt  is  not  subjected  to  working  stresses,  this  would  be  safe  for 
either  wrought  iron  or  mild  steel.  On  the  other  hand,  if  the 
final  turning  be  done  suddenly  by  means  of  a  jerking  motion  or 
a  blow,  or  a  long  wrench  be  used,  or  even  an  extra-strong  grad- 
ual pull  be  exerted,  there  is  evident  danger  of  /  having  a  value 
beyond  the  elastic  limit  of  the  material,  even  reaching  the  ulti- 
mate strength. 

It  will  be  noticed  also  that  the  torsional  action  increases  the 

fiber  stress  over  that  due  to  pure  tension  in  the  ratio  of  —  — ,  i.e., 

in  this  problem,  an  increase  of  over  17  per  cent.  In  general  this 
increase  will  be  from  15  to  20  per  cent,  depending  chiefly  upon 
the  relation  existing  between  /*  and  //.  It  should  also  be  noted 
that  the  pure  tension,  T,  induced  in  the  bolt  by  the  moment  PI 
may  be  taken  as  the  measure  of  the  pressure  existing  between 
the  surfaces  E-F  (Fig.  78).  In  our  problem  this  pressure,  for 
P=30  Ibs.,  would  become  30X74.467=2234  Ibs. 

As  a  general  rule  the  length  of  wrench  used  by  the  workman 
is  fifteen  or  sixteen  times  d\,  the  diameter  of  bolt,  and  it  may  be 
stated  that  T  =  P  for  U.  S.  standard  threads. 


MACHINE  DESIGN. 


92.  Calculation  of  Bolts   Subject  to  Elongation. — Next  con- 
sider the  case  shown  in  Fig.  85.     Suppose  that  the  nut  is  screwed 


up  with  a  resulting  tensile  stress  in  the  bolt 
=  T.  A  working  force  Q  tends  to  separate 
the  bodies  A  and  B  at  C-D.  Assume  that  Q 
acts  axially  along  the  bolt.  The  question  is, 
What  value  may  Q  have  without  opening  the 
joint  C-D? 

A    is  the  cross-sectional  area  of  the  bolt; 

L  is  the  original  length  between  bolt-head 
and  nut  when  A  and  B  are  just  in 
contact  at  C-D  but  not  compressed; 


rrri 


Fro.  85. 


TO  is  the  tensile  stress  in  bolt  due  to  screwing  up; 

A     is  the  total  elongation  of  bolt  due  to  TO  ; 

E   is  the  coefficient  of  elasticity  of  the  bolt  material. 


Then, 


since 


unit  strain 
unit  stress 


=-£,  it  follows  that 


(i) 


In  any  given  case  this  can  be  solved  for  A. 
Let  A'  be  the  area  of  A  and  B  compressed  by  the  bolt  action; 
A'  is  the  total  compression  (i.e.,  shortening)  of  A  and  B, 

due  to  the  tightening  of  the  bolt; 
Co  is  the  total  compressive  stress  which  produces  ^'; 

.*.  Co  =  TQ. 
E'  is  the  coefficient  of  elasticity  of  the  material  A ,  B.     Then 


—  =— 
Co  ~E'' 

A' 


(2) 


BOLTS  AND  SCREWS.  133 

This  can  be  solved  for  X . 

Now  consider  the  condition  when  a  working  force,  Q,  acts 
tending  to  elongate  the  bolt  so  that  A  and  B  will  just  be  ready 
to  separate  at  C-D.  In  order  that  this  separation  may  begin, 
the  bolt,  already  elongated  an  amount  A,  must  be  elongated  a 
further  amount  A'. 

For  incipient  separation  the  total  elongation  of  the  bolt  then 
=  A+A',  and  the  total  stress  in  the  bolt  corresponding  to  this 
elongation,  =  T',  can  be  determined  from  the  equation 


T 
A 


(3) 


Considering  the  bolt-head  as  a  free  body  (Fig.  86),  it  follows 

\Y^\]     that  the  forces  acting  on   it  at  any  instant  will  be,  C, 

£  I  ^     the  reaction  of  the   material  of  A  due  to  its  resistance 

•£'     to  compression;  Q,  the  working  force;  and  T,  the  ten- 

Fw.  86.   sion  in  the  bolt.     Hence 


(4) 


When  the  bolt  is  first  screwed  up,  and  Q=o,  then  C  =  T", 
and  T  =  JQ,  the  tension  due  to  screwing  up.  When  Q  comes 
into  action,  C  is  partly  relieved,  and  when  Q  reaches  such  a 
value  that  the  surfaces  are  about  to  part,  then  C=o  and  Q  = 
T  =  T'.  (See  equation  (3)  .) 

An  examination  of  these  formulae  shows  certain  facts  which 
may  be  stated  as  follows:  The  tightness  of  the  joint  C-D  de- 
pends upon  the  compressibility  of  A  and  B. 

Anything  which  increases  the  total  compression,  X',  increases 
the  tightness  of  the  joint.  This  may  be  accomplished  by  in- 
creasing L  or  Co,  or  decreasing  A'  .  It  may  also  be  increased 


134  MACHINE  DESIGN. 

by  the  introduction  of  a  highly  elastic  body  (i.e.,  gasket)  between 
A  and  B. 

It  also  follows  that  the  tension  in  the  bolt  when  the  joint  is 
about  to  open,  T',  must  be  greater  than  the  tension  due  to 
screwing  up,  TO,  and  therefore  if  Q  be  limited  to  a  value  equal 
to  or  less  than  T0,  there  will  be  no  opening  of  the  joint.  In 
general,  A'  is  large  compared  with  A,  and  X'  very  small  com- 
pared with  X,  so  that  T'  is  not  much  greater  than  TO.  In  order 
to  be  sure  of  a  tight  joint  the  initial  tension  should  be  taken 
T0  =  2(2. 

93.  Problem  I. — Calculate  the  bolts  for  a  "blank"  end  for 
a  6-inch  pipe  using  flanged  couplings  with  ground  joints,  and 
no  gaskets,  as  shown  in  Fig.  87.  The  excess  internal  pressure 
is  to  be  150  Ibs.  per  square  inch. 


FIG.  87. 
The  area  subjected  to  pressure  has  a  diameter  of  7^  ins.; 


hence  the  total  working  pressure  =  i  50  X 


=  6627  Ibs. 


The  number  of  bolts  is  determined  by  the  distance  they  may 
be  spaced  apart  without  danger  of  leakage  due  to  the  springing 
of  the  flange  between  the  bolts.  This  distance  may  be  taken 
equal  to  four  or  five  times  the  thickness  of  the  flange.  In  the 
problem  under  consideration,  the  diameter  of  the  bolt  circle  will 
be  approximately  9  ins.,  and  using  six  bolts,  the  chord  length 
between  consecutive  ones  will  be  about  4^  ins.,  which  is  per- 
fectly safe. 


BOLTS  AND  SCREWS.  135 

With  six  bolts 


Take  T0  =2Q  =  2210  Ibs. 

With  a  direct  tension  of  2210  Ibs.  due  to  screwing  up,  there 
is  also  the  stress  due  to  torsion.  As  stated  in  §  91,  this  may 
increase  the  fiber  stress  20  per  cent  over  that  due  to  direct  ten- 
sion. To  allow  for  this  the  bolts  used  must  be  capable  of  safely 
sustaining  a  stress  of  2210X1.20  =  2650  Ibs. 

The  allowable  unit  stress  here  may  be  taken  rather  high, 
since  the  conditions  after  once  screwing  up  approximate  a  steady 
load.  Assume  steel  bolts  with  an  allowable  unit  stress  of 
15,000  Ibs. 

The  area  of  each  bolt  at  the  bottom  of  the  thread  will  then 

26^0 
be  -  =0.177  sq.  in.     This  value  lies  between  a  .A-hich  and 

a  f-inch  bolt.  Select  the  latter  with  an  area  of  0.202  sq.  in.  To 
exert  an  initial  tension  of  2250  Ibs.  in  a  f-inch  bolt  would  re- 
quire a  pull  of  about  30  Ibs.  on  a  lo-inch  wrench.  (See  §  91.) 
These  values  just  about  correspond  to  actual  conditions  in  prac- 
tice. 

94.  Problem  II.  —  It  is  required  to  design  the  fastenings  to 
hold  on  the  steam-chest  cover  of  a  steam-engine.  The  opening 
to  be  covered  is  rectangular,  io"Xi2".  The  maximum  steam- 
pressure  is  100  Ibs.  per  square  inch.  The  joint  must  be  held 
steam-tight.  Studs  of  machinery  steel  having  an  ultimate  ten- 
sile strength  of  60,000  Ibs.  per  square  inch  will  be  used. 

The  total  working  pressure  =  zoX  12  X  100=  12,000  Ibs. 

The  number  of  studs  to  be  used  will  be  governed  by  the  dis- 
tance they  may  be  spaced  without  springing  of  the  cover.  The 
thickness  of  the  latter  being  assumed  to  be  f  inch  at  the  edge, 
the  spacing  should  not  exceed  5  Xf"  =  J^",  say  4  inches. 


MACHINE  DESIGN. 


_Q 


The  opening  is  io"X  12",  as  shown  in  Fig.  88.  There  must 
be  a  band  about  f  or  f  inch  wide 
around  this  for  making  the  joint, 
upon  which  the  studs  must  not  en- 
croach. This  makes  the  distance 
between  the  centers  of  the  vertical 
rows  of-  studs  about  14  inches,  and 
between  the  horizontal  rows  about 
12  inches.  Twelve  studs  can  be  used 
if  arranged  as  shown  in  the  figure. 
The  greatest  distance,  that  between 
the  studs  across  the  corners,  will  but 
slightly  exceed  the  allowable  4  inches. 

With  12  studs,  the  working  load  on  each  =  Q  = 

Ibs. 


12000 


1000  Ibs. 


Allowing  20  per  cent  for  torsional  stress,  increase  this  to 
2400  Ibs. 

Allowing  a  unit  stress  of  15,000  Ibs.,  as  in  Problem  I,  we 

have  as  the  area  of  the  stud  at  the  bottom  of  thread  —   —  =  o.  160. 

15000 

This  corresponds  to  a  A-inch  stud.  Since  a  workman  may 
readily  stress  a  bolt  of  this  size  beyond  the  elastic  limit  by  exert- 
ing too  great  a  pull  in  tightening,  many  designers  would  increase 
these  studs  to  f  inch  or  even  f  inch. 

95.  Design  of  Bolts  for  Shock.  —  The  elongation  of  a  bolt 
with  a  given  total  stress  depends  upon  the  LENGTH  and  AREA 
of  its  least  cross-section.  Suppose,  to  illustrate,  that  the  bolt, 
Fig.  89,  has  a  reduced  section  over  a  length  /  as  shown.  This 
portion,  A,  has  less  cross-sectional  area  than  the  rest  of  the  bolt, 
and  when  any  tensile  force  is  applied,  the  resulting  UNIT  stress 
will  be  greater  in  A  than  elsewhere.  The  unit  strain,  or  elonga- 
tion, will  be  proportionately  greater  up  to  the  elastic  limit;  and 


BOLTS  AND  SCREWS. 


137 


if  the  elastic  limit  is  exceeded  in  the  portion  A,  the  elongation 
there  will  be  far  greater  than  elsewhere.  If  there  is  much  differ- 
ence of  area  and  the  bolt  is  tested  to  rup- 
ture, the  elongation  will  be  chiefly  at  A. 
There  would  be  a  certain  elongation  PER 
INCH  of  A  at  rupture.  Hence  the  greater 
the  length  of  A,  the  greater  the  total  elonga- 
tion of  the  bolt.  If  the  bolt  had  not  been 
reduced  at  A,  the  minimum  section  would 
be  at  the  root  of  the  screw-threads.  The 
FIG.  89.  FIG.  90.  axiai  lcngth  of  this  section  is  very  small. 
Hence  the  elongation  at  rupture  would  be  small.  Suppose  there 
are  two  bolts,  A  with  and  B  without  the  reduced  section.  They 
are  alike  in  other  respects.  They  are  subjected  to  equal  tensile 
shocks.  Let  the  energy  of  the  shock  =  E.  This  energy  is  di- 
vided into  force  and  space  factors  by  the  resistance  of  the  bolts. 
The  space  factor  equals  the  elongation  of  the  bolt.  This  is 
greater  in  A  than  in  B,  because  of  the  yielding  of  the  reduced 
section.  But  the  product  of  force  and  space  factors  is  the  same 
in  both  bolts,  =£;  hence  the  resulting  stress  in  the  minimum 
section  is  less  for  A  than  for  B.  The  stress  in  A  may  be  less 
than  the  breaking  stress,  while  the  greater  stress  in  B  may 
break  it.  THE  CAPACITY  OF  THE  BOLT  TO  RESIST  SHOCK  is 

THEREFORE     INCREASED     BY     LENGTHENING     ITS     MINIMUM     SEC- 
TION TO  INCREASE  THE  YIELDING  AND  REDUCE  STRESS.      This  IS 

not  only  true  of  bolts,  but  of  all  stress  members  in  machines. 

The  whole  body  of  the  bolt  might  have  been  reduced,  as 
shown  by  the  dotted  lines  in  Fig.  89,  with  resulting  increase  of 
capacity  to  resist  shock.  Turning  down  a  bolt,  however,  weak- 
ens it  to  resist  torsion  and  flexure,  because  it  takes  off  the  material 
which  is  most  effective  in  producing  large  polar  and  rectangular 
moments  of  inertia  of  cross-section.  If  the  cross-sectional  area 
is  reduced  by  drilling  a  hole,  as  shown  in  Fig.  90,  the  torsional 


138  MACHINE  DESIGN. 

and  transverse  strength  is  but  slightly  decreased,  but  the  elon- 
gation will  be  as  great  with  the  same  area  as  if  the  area  had 
been  reduced  by  turning  down. 

Professor  Sweet  had  a  set  of  bolts  prepared  for  special  test. 
The  bolts  were  i^  inches  diameter  and  about  12  inches  long. 
They  were  made  of  high-grade  wrought  iron,  and  were  dupli- 
cates of  the  bolts  used  at  the  crank  end  of  the  connecting-rod 
of  one  of  the  standard  sizes  of  the  Straight-line  Engine.  Half 
of  the  bolts  were  left  solid,  while  the  other  half  were  carefully 
drilled  to  give  them  uniform  cross-sectional  area  throughout. 
The  tests  were  made  under  the  direction  of  Professor  Carpenter 
at  the  Sibley  College  Laboratory.  One  pair  of  bolts  was  tested 
to  rupture  by  tensile  force  gradually  applied.  The  undrilled 
bolt  broke  in  the  thread  with  a  total  elongation  of  0.25  inch. 
The  drilled  bolt  broke  between  the  thread  and  the  bolt-head 
with  a  total  elongation  of  2.25  inches.  If  it  be  assumed  that 
the  mean  force  applied  was  the  same  in  both  cases,  it  follows 
that  the  total  resilience  of  the  drilled  bolt  was  nine  times  as  great 
as  that  of  the  solid  one.  ''Drop  tests,"  i.e.,  tests  which  brought 
tensile  shock  to  bear  upon  the  bolts,  were  made  on  other  similar 
pairs  of  bolts,  which  tended  to  confirm  the  general  conclusion. 

96.  Problem. — It  is  required  to  design  proper  fastenings  for 
holding  on  the  cap  of  a  connecting-rod  like  that   shown  in  Fig. 
91.    These  fastenings  are  required  to  sus- 
tain shocks,  and  may  be  subjected  to  a 
maximum  accidental  stress  of  20,000  Ibs. 
There  are  two  fastenings,  and  therefore 
each  must  be  capable  of  sustaining  safely 
a  stress  of  10,000  Ibs.     They  should  be  *MIN  ^ — - — '  ;\\\\ 
designed  to  yield  as  much  as  is  consistent 
with  strength ;  in  other  words,  they  should 
be  tensile  springs  to  cushion  shocks,  and 
thereby  reduce  the  resulting  force  they  have  to  sustain.     Bolts 


BOLTS  AND  SCRbWS.  139 

should  therefore  be  used,  and  the  weakest  section  should  be. 
made  as  long  as  possible.  Wrought  iron  will  be  used  whose 
tensile  strength  is  50,000  Ibs.  per  square  inch.  The  stress  given 
is  the  maximum  accidental  stress,  and  is  four  times  the  working 
stress.  It  is  not,  therefore,  necessary  to  give  the  bolts  great 
excess  of  strength  over  that  necessary  to  resist  actual  rupture 
by  the  accidental  force.  Let  the  factor  of  safety  be  2.  This 
will  keep  the  maximum  fiber  stress  within  the  elastic  limit. 
Then  the  cross-sectional  area  of  each  bolt  must  be  such  that 
it  will  just  sustain  10,000X2=20,000  Ibs.  This  area  is  equal 
to  20,000-^50,000=0.4  sq.  in.  This  area  corresponds  to  a 
diameter  of  0.71  inch,  and  that  is  nearly  the  diameter  of  a 
|-inch  bolt  at  the  bottom  of  the  thread;  hence  f-inch  bolts  will 
be  used.  The  cross-sectional  area  of  the  body  of  the  bolt 
must  now  be  made  at  least  as  small  as  that  at  the  bottom  of  the 
thread.  This  may  be  accomplished  by  drilling. 

97.  Jam-nuts. — When  bolts  are  subjected  to  constant  vibra- 
tion there  is  a  tendency  for  the  nuts  to  loosen.  There  are  many 
ways  to  prevent  this,  but  the  most  common  one  is  by  the  use  of 
jam-nuts.  Two  nuts  are  screwed  on  the  bolt;  the  under  one 
is  set  up  against  the  surface  of  the  part  to  be  held  in  place,  and 
then  while  this  nut  is  held  with  a  wrench  the  other  nut  is  screwed 
up  against  it  tightly..  Suppose  that  the  bolt  has  its  axis  vertical 
and  that  the  nuts  are  screwed  on  the  upper  end.  The  nuts  being 
screwed  against  each  other,  the  upper  one  has  its  internal  screw 
surfaces  forced  against  the  under  screw  surfaces  of  the  bolt,  and 
if  there  is  any  lost  motion,  as  there  almost  always  is,  there  will 
be  no  contact  between  the  upper  surfaces  of  the  screw  on  the 
bolt  and  the  threads  of  the  nut.  Just  the  reverse  is  true  of  the 
under  nut;  i.e.,  there  is  no  contact  between  the  under  surfaces 
of  the  threads  on  the  bolt  and  the  threads  on  the  nut.  There- 
fore no  pressure  that  comes  from  the  under  side  of  the  under 
nut  can  be  communicated  to  the  bolt  through  the  under  nut 


14°  MACHINE  DESIGN. 

directly,  but  it  must  be  received  by  the  upper  nut  and  com- 
municated by  it  to  the  bolt,  since  it  is  the  upper  nut  alone  that 
has  contact  with  the  under  surfaces  of  the  thread.  Therefore 
the  jam-nut,  which  is  usually  made  about  half  as  thick  as  the 
other,  should  always  be  put  on  next  to  the  surface  of  the  piece 
to  be  held  in  place. 

98.  Calculation  of  Screws  for  Transmission  of  Power.  — 
(c}  Screws  are  frequently  used  to  transmit  power.  A  screw- 
press  is  a  common  example,  while  the  action  of  spiral  gears, 
including  worms  and  wheels,  is  that  of  screws  and  subject  to 
the  same  analysis. 

The  general  action  of  screw  and  nut  has  already  been  treated. 
(See  §  87.) 

With  a  square-thread  screw  it  was  found  that  the    moment 


Pl=M,  required  to  raise  a  load  W,  will  be=Wr  (5) 

27tr-pfJL    V- 

(page  124). 

This  will  induce  a  fiber  stress  /«  =~T~  (see  equation  (19)  ), 

W 
which  must  be  combined  with  the  tension,  }t  =—  ,  in  order  to 

^T. 

get  the  actual  unit  stress,  /,  remembering 

-'  .....     (21) 


Let  «=number  of    complete    thread    surfaces    in    contact 
between    the   nut  and  screw,  and    the    projected    area    equals 


n-(dl2-d22)  to  bear  the  load  W. 
4 


where  K  is  the  allowable  pressure  per  square  inch  of  projected 
thread  area. 


BOLTS  AND  SCREWS. 


141 


For  nuts  and  bolts  which  are  used  as  fastenings  we  may  take: 
#  =  2500  Ibs.  for  wrought  or   cast   iron  running  on  the  same 

material  or  on  bronze ;  m 

K  =  3000  Ibs.  for  steel  on  steel  or  bronze. 

With  good  lubrication,  where  the  screw  and  nut  are  used  to 
transmit  power,  we  may  take  the  values  given  in  the  following 

table : 

TABLE  VII. 


Rubbing  Speed  in  Feet 
per  Minute. 

Value  of  K. 

Iron. 

Steel. 

2500 
1250 
850 
400 

200 

3000 
1500 

1000 

500 
250 

The  value  of  //  has  been  experimentally  determined  by  Pro- 
fessor Kingsbury.*  He  concludes  that  for  metallic  screws  turn- 
ing at  extremely  slow  speeds,  under  any  pressure  up  to  14,0x30 
Ibs.  per  square  inch  of  bearing  surface,  and  freely  lubricated 
before  application  of  the  pressure,  the  following  coefficients  of 
friction  may.  be  used. 

TABLE  VIII. 


Lubricant. 

/'• 

Lard-oil  
Heavy-machinery  oil  (mineral)  
Heavy-machinery  oil  and  graphite  in  equal 

0.  II 

0.143 

Regarding  the  efficiency  of  the  square-screw  thread  to  trans- 
mit energy,  we  may  reason  as  follows: 

useful  work 
The  efficiency  =  ,= 


*  Transactions  A.  S.  M.  E.,  Vol.  XVII,  pp.  96-116. 


142  MACHINE  DESIGN. 

Wp 


which  becomes  for  one  turn 


Wp  _      2xr  _     W  tan  a  tan  a 

=    H    =  W  tan  (a  +  $  =  tan  (a 


(22) 


From  this  it  appears  that  e  becomes  o,  for  a=o°  and  for 
a  =90°  —  <£,  and  must  therefore  have  a  maximum  value  between 
these  limits.  To  determine  this  maximum,  write 

tan  a 

-JT  =tan  a  cot  (a  +  <f>). 


tan  (oH  , . 

Taking  the  first  differential  and  equating  to  o, 

cot  (a  +  <i)  tan  a 

o ~  •  o  / — r~r\  =o- 

cos2  a         sin2  (a  +  (f>) 

Solving  which  gives 

«=45°-- 

To  find  the  corresponding  value  of  e,  write  (from  (22)  ) 
tan(45°-f)        «an(45--f) 


To  lower  W  with  a  square-threaded  screw, 


[The  value  of  /,  can  be  found  from  this,  as  explained  in  the 
section  on  raising  W,  and  combined  with  ft  to  obtain  /.] 


BOLTS  AND  SCREWS.  143 

Regarding  the  efficiency  in  this  case, 

if  a<  $  the  load  will  not  sink  of  itself  (i.e.,  overhaul), 

if  a  =  <f>  we  have  a  condition  of  equilibrium, 

if  a  >0  the  load  will  sink  of  itself  (i.e.,  overhaul). 

For  a  screw  which  will  not  overhaul  it  becomes  evident  that 
the  limiting  value  of  a  is  </>  and  the  maximum  efficiency 

tan  a  tan  <f>      i—  tan2  2<£ 


The  efficiency  of  a  screw  which  will  not  overhaul  can  there- 
fore never  exceed  0.5  or  50  per  cent. 

For  V  threads,  with  /?  =  angle  of  V,  with  a  plane  normal  to 
the  axis  of  the  screw  for  raising  load, 


which  is  evidently  greater  than  (5),  and 

tan  a(i  —  p.  tan  a.  sec 


which  is  evidently  less  than  e  for  square  threads.  (See  equa- 
tion (22).) 

It  is  clear,  then,  that  square  threads  should  be  used  in  pref- 
erence to  V  threads  for  screws  for  power  transmission. 

For  lowering  the  load  with  V  threads 


2nr  -\-pfjiSGC 


(26) 


99.  Problem.  —  Design  a  screw  to  raise  20,000   Ibs.     The 
screw  must  not  overhaul. 

What  moment  need  be  exerted  to  lift  the  load  ? 


144  MACHINE  DESIGN. 

What  will  be  the  efficiency  of  the  screw  ? 
Select  a  square-thread  screw  of  machinery  steel  running  in 
a  bronze  nut. 

For  a  screw  which  will  not  overhaul  a  must  be  less  than  <£. 

.'.  tan  a<  //. 

To  be  safe  against  overhauling  with  the  materials  used  and 
good  lubrication,  /*  must  not  be  given  a  greater  value  than  o.io. 

.'.  a<5°45'     and     ^  =  5°  45'. 

The  pure  tension  =  20,000  lbs.  =  PF.     In  the  preliminary  calcu- 
lations, to  allow  for  the  effect  of  torsion,  this  will  be  increased 

215000 
so  that  /=      ,     . 

A 

In  this  equation  /  is  the  allowable  unit  stress  in  pounds  per 
square  inch,  and  A  is  the  area  of  the  screw  at  the  bottom  of  the 
thread  in  square  inches. 

Assume  that  this  screw  is  frequently  loaded  and  unloaded, 
and  not  subject  to  shocks  nor  reversal  of  stress  so  that  /  = 

12,000  Ibs.  per  square  inch  for  mild  steel.     Then  A  = = 

12000 

2.08  sq.  ins.     This  corresponds  to  a  diameter  of  if  inches  at 
bottom  of  thread. 

P 

From  tan  a  =  -*— ,  we  have 

27rr 

p  =  2nr  tan  a. 

Remembering  that  for  square  threads  the  depth  of  the  thread 
=  — ,  and  that  r^  is  the  radius  at  the  bottom  of  thread,  and 


it  follows  that 

4 


BOLTS  4ND  SCREWS.  145 


7          P\ 
=  2n\r2  +  -)  tan  a, 


27r  tan  a 
p — p  =  2nr2  tana, 

4 


27Tf2  tan  a 
*-—  J-  -• 

i  —  tan  a 

2 

2X71X0.8125X0.1 

.'.  £  =  -  -  =  .606  inch. 

1-1.57X0.1 

This  is  not  a  thread  to  be  easily  cut  in  the  lathe.  It  would 
be  desirable  to  modify  the  value  of  p  so  that  the  thread  can  be 
readily  cut.  It  is  obvious  that  p  cannot  be  increased  without 
increasing  r  proportionately,  else  a  will  have  a  greater  value 
than  is  allowable.  It  will  be  more  economical  to  reduce  p. 
The  nearest  even  value  would  be  J  inch,  and  this  will  be  selected. 
Check  this  for  strength: 

d2  =  1.62  5  inches,       r2  =0.8125  inch,  />  =  0,5  inch. 

d\  =  2.  125  inches,       r\  =  1.0625  inches, 
d=  1.875  inches,         ^=°-9375  inch, 

-h  o  ^ 

tana=  —  =  -  -  --  =  0.087, 
27ZT     2X71X0.9375 

which  is  safe,  as  it  is  less  than  the  value  of  /*  =  o.io. 
From  equation  (5),  the  moment, 


0.5  +  2X71X0.0375X0.10 
( 

3496  in.-lbs. 


146  MACHINE  DESIGN. 

From  equation  (19),  the  fiber  stress  due  to  torsion 
cPl    2PI 


lbs<  per  Square 
From  equation  (17),  the  unit  stress  due  to  tension, 

W     20000  . 

/<=^  =  ^j-=9597  «*•  per  square  inch. 

From  equation  (21),  the  combined  stress, 


=  12,379  Ibs.  per  square  inch, 

which  is  near  enough  12,000  to  be  considered  safe. 
The  efficiency,  from  equation  (22), 


tan  a 
! 


(a-f  <£)* 

Since  tan  a  =0.087,        a  =  5°- 
Since  n  =  tan  <£  =  o.  10,    <£  =  5°  45'. 


/,     tan  10°  45'  =0.1899. 


The  height  of  the  nut  is  determined  from  the  equation 


BOLTS  AND  SCREWS.  147 

in  which  W  is  the  load,  n  the  number  of  complete  threads  in 
the  nut,  d\  the  outside  and  d2  the  inside  diameter  of  thread. 
K  is  the  allowable  pressure  in  pounds  per  square  inch,  and  its 
value  depends  upon  the  speed.  See  table  in  sec.  98. 

Assuming  the  screw  to  have  a  rubbing  velocity  of  less  than 
50  feet  per  minute,  K  =  3000.     Then 

W  20000 


-  1.6252) 


=  4.5,  nearly. 
The  height  of  nut  = 


CHAPTER  IX. 

MEANS   FOR   PREVENTING   RELATIVE   ROTATION. 

100.  Classification  of  Keys.  —  Keys  are  chiefly  used  to  pre- 
vent relative  rotation  between  shafts  and  the  pulleys,  gears,  etc., 
which  they  support.     Keys  may  be  divided  into  parallel  keys, 
taper  keys,  disk  keys,  and  feathers  or  splines. 

101.  Parallel  Keys.  —  For  a  PARALLEL  KEY  the  "seat,"  both 
in  the  shaft  and  the  attached  part,  has  parallel  sides,  and  the 
key  simply  prevents  relative  rotary  motion.     Motion  parallel  to 
the  axis  of  the  shaft  must  be  prevented  by  some  other  means, 
as  by  set-screws  which  bear  upon  the  top  surface  of  the  key,  as 
shown  in  Fig.  92.     A  parallel  key  should  fit  accurately  on  the 
sides  and  loosely  at  the  top  and  bottom.     The  following  table 
(IX)  for  dimensions  for  parallel  keys  is  from  Richards's  "Manual 
of  Machine  Construction." 

TABLE  IX. 


ij        i|        if        2          2}       3          3i 

&&&MMHA 
}         £        |          &        J         &        f 


Diameter  of  shaft  =  d  =  i 
Width  of  key        =  ™= 
Height  of  key       =  *  =  . 

Excellent  parallel  keys  are  made  from  cold-rolled  steel  with- 
out need  of  any  machining. 

102.  Taper  Keys.  —  A  TAPER  KEY  has  parallel  sides  and  has 
its  top  and  bottom  surfaces  tapered,  and  is  made  to  fit  on  all 
four  surfaces,  being  driven  tightly  "home."  It  prevents  rela- 
tive motion  of  any  kind  between  the  parts  connected.  If  a  key 
of  this  kind  has  a  head,  as  shown  in  Fig.  93,  it  is  called  a  "draw 
key,"  because  it  is  drawn  out  when  necessary  by  driving  a 

148 


MEANS  FOR  PREVENTING  RELATIVE  ROTATION.  149 

wedge  between  the  hub  of  the  attached  part  and  the  head  of  the 
key.  When  a  taper  key  has  no  head  it  is  removed  by  driving 
against  the  point  with  a  "key-drift." 


FIG.  92.  FIG.  93.  FIG.  94. 

d 

John  Richards's  rule  for  keys  is  (see  Fig.  94)  «;=-.     /  has 

such  value  that  a  =30°.  This  rule  is  deviated  from  somewhat, 
as  shown  by  the  following  table  (X)  taken  from  Richards's 
"Manual  of  Machine  Construction,"  page  58: 

TABLE  X. 

<Z-i        1}      z|       if      a         2\      3         3i  4         5  6  7  8 

»-  1       ft      I         A       \         f        f         I  i          it  if  i*  if 

'-  ft     ft       i         ft      ft       f        ft        I        i         »      ft  f  i 

When  two  or  more  keys  are  used,  -w=d^-6,  t  being,  as  before, 
of  such  value  that  a  shall  =  30°. 

The  taper  of  keys  varies  from  \  to  £  inch  to  the  foot. 

103.  Fitting  Shaft  and  Hub. — In  using  taper  keys  it  is  cus- 
tomary to  bore  out  the  hub  slightly  larger  than  the  diameter  of 
the  shaft  so  that  the  wheel  may  be  readily  removed  after  the 
key  is  withdrawn.  This  allowance  in  diameter  should  not  be 
greater  than  that  for  a  running  fit,  say, 


in  which   formula  A  is  the  difference  in  diameter  between  the 
bore  of  hub  and  size  of  shaft,  expressed  in  decimal  parts  of  an 


MACHINE  DESIGN. 


inch,  and  D  is  the  nominal  diameter  of  shaft  in  inches.  Where 
the  parts  do  not  have  to  be  taken  apart  frepuently,  it  is  vastly 
better  to  use  a  driving  fit,  i.e.,  to  bore  out  the  hub  smaller  than 

D 


the  diameter  of  the  shaft  by  an  amount  A  = 


+  0.5 


1000 


and  to  use 


parallel  keys.  » 

Where  a  single  taper  key  is  used  the  effect  is  to  make  the 
wheel  and  shaft  eccentric,  as  can  be  seen  in  Fig.  95.  The  bear- 
ing is  limited  to  two  points,  A,  B,  and  the  connection  is  unstable 
for  the  transmission  of  power. 


FIG.  95. 


FIG.  96. 


If  great  care  is  not  exercised  in  having  the  taper  of  keyway 
exactly  the  same  as  the  taper  of  the  keys,  a  further  difficulty 
arises  in  that  the  wheel  will  be  canted  out  of  a  true  normal  plane 
to  the  shaft-axis.  This  can  be  seen  in  Fig.  96. 

By  using  two  keys,  placed  a  quarter  or  third  of  the  circum- 
ference apart,  a  much  more  stable  connection  is  obtained,  as  it 
will  give  three  points  of  bearing,  A,  B,  and  C.  {See  Fig.  97.) 
Eccentricity  is  not  avoided  by  this  method. 

Another  method  is  that  shown  in  Fig.  98.  Here  the  hub  is 
bored  out  somewhat  larger  than  the  shaft  and  a  bushing  B  is 
used  opposite  the  key.  This  bushing  has  its  outside  diameter 
exactly  the  diameter  of  the  bore,  and  its  inner  diameter  exactly 


MEANS  FOR  PRESENTING  RELATIVE  ROTATION. 


the  diameter  of  the  shaft,  and  it  surrounds  slightly  less  than 
half  the  shaft.    When  the  key  is  driven  home  this  insures  the 


FIG.  97. 


FIG.  98. 


wheel  and  shaft  being  concentric,  but  extra  care  must  be  taken 
that  key  and  key  way  have  precisely  the  same  taper.  Three 
keys  uniformly  spaced  around  the  shaft  also  give  a  means  of 
making  the  wheel  and  shaft  concentric.  This  method  is  rarely 
used. 

104.  Woodruff  Keys. — The  Woodruff  or  disk  system  of  keys 
is  used  by  some  manufacturers.  The  key  is  a  half  disk,  as  can 
be  seen  in  Fig.  99.  Under  this  system  the  keyway  is  cut  lon- 
gitudinally in  the  shaft  by  means  of  a  milling-cutter.  This  cut- 
ter corresponds  in  thickness  to  the  key  to  be  inserted,  and  is  of 


FIG.  99. 


FIG.  100. 


a  diameter  corresponding  to  the  length  of  the  key.  The  key 
being  semicircular,  it  is  sunk  into  the  shaft  as  far  as  will  allow 
sufficient  projection  of  the  key  above  the  surface  to  engage  the 
keyway  in  the  hub. 


152 


MACHINE   DESIGN. 


Owing  to  its  peculiar  shape  the  key  may  be  slightly  inclined, 
so  that  it  will  serve  to  support  the  wheel  on  a  vertical  shaft,  pro- 
vided the  key-seat  in  the  hub  is  made  tapering  and  of  the  proper 
depth. 

104.  Saddle,  Flat  and  Angle  Keys. — Saddle  keys  (Fig.  100,  A) 
and  keys  on  flats  (Fig.  100,  B)  are  used  occasionally.  They 
have  not  the  holding  power  of  sunk  keys. 


FIG.  ioo. 

The  type  of  key  shown  in  Fig.  ioo  C  has  much  to  be  said  in 
its  favor  both  as  regards  ease  and  accuracy  in  obtaining  a  stable 
connection  and  also  as  regards  suitability  of  form  to  resist  stresses. 
It  will  be  noted  that  the  surfaces  are  normal  to  the  lines  of  action 
of  the  forces  transmitted.  The  pressure  per  square  inch  should 
not  exceed  17,000  pounds.  The  height  of  key  is  taken  equal 
to  0.2  diameter  of  shaft. 

The  Kernoul  key,  shown  in  Fig.  ioo  D,  is  for  use  in  driving 
in  only  one  direction.    A  portion  of  the  hub  is  cut  out  so  as  to 


MEANS  FOR    PREVENTING  RELATIVE  RELATION.  153 

form  an  eccentric  slot.  In  this  the  key  fits  as  shown.  The 
inner  face  of  the  key,  curved  to  the  radius  of  the  shaft,  should 
be  left  rough  so  as  to  seize  the  shaft,  while  the  outer  face,  curved 
to  fit  the  slot  in  the  hub,  is  smooth  finished.  When  the  shaft 
rotates  (in  this  case)  counter-clockwise,  the  resistance  to  the 
hub's  motion  being  then  as  indicated  by  the  arrow,  the  surface 
of  the  slot  tends  to  slide  up  on  the  key,  causing  it  to  wedge  in 
between  the  shaft  and  hub,  forming  a  firm  connection.  When 
the  shaft  rotates  in  the  opposite  direction  and  the  resistance 
to  the  hub's  motion  is  reversed,  the  slot  of  the  hub  tends  to  leave 
the  key,  relieving  the  pressure  and  permitting  easy  removal  from 
the  shaft.  At  "  a  "  and  "b  "  there  are  counter-sunk  screws  for 
setting  up  and  loosening  the  key.  In  Fig.  100  E  is  shown  a  special 
form  of  this  type  of  key.  It  is  known  as  the  Barbour  key  and 
is  chiefly  used  for  fastening  the  cams  on  the  shafts  of  stamp 
mills* 

In  the  study  of  keys  which  drive  in  one  direction  only  it  is 
proper  to  include  the  roller  ratchet.  The  simplest  form  is  shown 
in  Fig.  100  F.  The  hub  is  recessed  as  shown  and  the  roller  R 
placed  in  the  recess,  held  in  position  by  a  spring.  The  direction 
of  shaft  rotation  and  hub  resistance  being  as  shown  the  roller 
becomes  wedged  between  the  two,  forming  a  driving  connection. 
With  reversal  of  direction  the  roller  is  freed  and  the  shaft  and 
hub  may  have  relative  motion. 

Generally  more  than  one  roller  is  used  and  the  mechanism 
takes  the  form  shown  in  Fig.  100  G.  A  is  a  hardened  and 
ground  steel  ring  or  bushing  and  B  should  also  be  hardened. 
Each  roller  should  be  held  in  place  by  a  spring  as  shown  at  C. 
Such  ratchets  permit  of  rapid  reciprocation.  Complete  details 
and  descriptions  of  further  clutches  of  this  type  will  be  found  in 
the  American  Machinist  of  Dec.  21,  1905. 

*  Patent  held  for  this  purpose  by  the  Risdon  Iron  Works.  The  distinguishing 
feature  from  the  plain  Kernoul  key  lies  in  the  use  of  the  inside  projecting  tongue 
which  fits  in  a  keyway  cut  in  the  shaft. 


154  MACHINE  DESIGN. 

105.  Strength  of  Sunk  Keys. — The  strength  of  the  latter  is 
the  measure  of  their  holding  power.     A  key  of  width  =w,  thick- 
ness ==/,  length  =/,  unit  shearing  strength  =/„  and  unit  crushing 
resistance  = /c  will  have  a  shearing  strength  =  jawl  and  a  crush- 
ing resistance  ]c\tl. 

If  r  =  radius  of  the  shaft,  the  moment  which  the  key  can 
resist  will  be  measured  by  rwlf8  or  \rtl]c^  whichever  is  smaller 
in  value. 

All  dimensions  being  expressed  in  inches  and  resistance 
in  pounds  per  square  inch,  the  moments  will,  of  course,  be 
expressed  in  inch-pounds.  Experiments  made  by  Professor 
Lanza  indicate  that  the  ultimate  value  for  }8  for  cast  iron  = 
30,000  Ibs.,  for  wrought  iron  =  40,000  Ibs.,  and  for  machinery 
steel  =  60,000  Ibs.  A  factor  of  safety  of  2  would  be  advis- 
able with  these  values. 

106.  Feathers  or  Splines  are  keys  that  prevent  relative  rota- 
tion, but  purposely  allow  axial  motion.     They  are  sometimes 


FIG.  101.  FIG.  102. 

made  fast  in  the  shaft,  as  in  Fig.  101,  and  there  is  a  key  "way" 
in  the  attached  part  that  slides  along  the  shaft.  Sometimes  the 
feather  is  fastened  in  the  hub  of .  the  attached  part,  as  shown 
in  Fig.  102,  and  slides  in  a  long  keyway  in  the  shaft. 

It  is  frequently  undesirable  to  have  the  feather  loose.  In 
such  cases  it  is  common  to  use  tit-keys  as  shown  in  Fig.  103. 
The  keys  may  be  fastened  to  either  hub  or  shaft.  The  tits 
are  forged  on  the  keys.  Corresponding  holes  are  drilled 
and  countersunk  in  the  piece  to  which  the  key  is  to  be  fastened, 


MEANS  FOR  PREVENTING  RELATIVE  ROTATION. 


'55 


and  after  the  key  is  placed  in  position  the  ends  of  the  tits  are 
riveted  over  to  hold  it  securely  in  place. 

Machine  screws  are  sometimes  used  in  place  of  tits,  but 
they  suffer  from  the  disadvantage  of  jarring  loose. 

A  satisfactory  way  of  holding  a  key  in  a  hub  is  shown  in 
Fig.  104. 

Where  the  end  of  a  stud  is  to  receive  change-gears  a  con- 
venient form  of  key  is  the  dovetail  shown  in  Fig.  105  in  cross- 


FIG.  103. 


FIG.  104. 


FIG.  105. 


section.     The  dovetailed  key-seat  is  generally  cut  with  a  mill- 
ing-cutter, and  is  made  a  tight  fit  for  the  key.     After  the  latter 
is  in  place  the  shaft  is  calked  against  it  at  A- A. 
For  feathers,  Richards  gives: 


rf-ij 

w=  i 

«-  i 


TABLE  XL 
2}        2* 


107.  Round  Taper  Keys.  —  For  keying  hand-wheels  and 
other  parts  that  are  not  subjected  to  very  great  stress,  a  cheap 
and  satisfactory  method  is  to  use  a  round  taper  key  driven 
into  a  hole  drilled  in  the  joint,  as  in  Fig.  106.  If  the  two  parts 
are  of  different  material,  one  much  harder  than  the  other, 
this  method  should  not  be  used,  as  it  is  almost  impossible  in 
such  case  to  make  the  drill  follow  the  joint.  For  these  keys 


MACHINE   DESIGN. 


it  is  customary  to  use  Morse  standard  tapers,  as  reamers  are 
then  readily  obtainable. 

108.  A  Cotter  is  a  key  that  is  used  to  attach  parts  that  are 
subjected  to  a  force  of  tension  or  compression  tending  to  sepa- 
rate them.  Thus  piston-rods  are  often  connected  to  both  pis- 
ton and  cross-head  in  this  way.  Also  the  sections  of  long 
pump-rods,  etc. 

Fig.  107  shows  machine  parts  held  against  tension  by  cot- 


FIG.  106. 


FIG.  107. 


ters.  It  is  seen  that  the  joint  may  yield  by  shearing  the  cot- 
ter at  AB  and  CD,  or  by  shearing  CPQ  and  ARS;  by  shear- 
ing on  the  surfaces  MO  and  LN;  or  by  tensile  rupture  of  the 
rod  on  a  horizontal  section  at  LM.  AD  of  these  sections  should 
be  sufficiently  large  to  resist  the  maximum  stress  safely.  The 
difficulty  is  usually  to  get  LM  strong  enough  in  tension;  but 
this  may  usually  be  accomplished  by  making  the  rod  larger 
or  the  cotter  thinner  and  wider.  It  is  found  that  taper  sur- 
faces if  they  be  smooth  and  somewhat  oily  will  just  cease  to 
stick  together  when  the  taper  equals  1.5  inches  per  foot.  The 
taper  of  the  rod  in  Fig.  107  should  be  about  this  value  in  order 
that  it  may  be  removed  conveniently  when  necessary 

From  consideration  of  the  laws  of  friction  it  is  obvious  that 
where  a  taper  cotter  is  used,  either  alone  as  in  Fig.  108  or  in 
connection  with  a  gib  as  in  Fig.  109,  the  angle  of  taper  a  must 
not  exceed  the  friction  angle  (f>.  That  is,  if  the  coefficient  of 
friction  be  /*,  then  /i=tan  0  and  tan  a  must  be  less  than  tan  (f> 


MEANS  FOR  PRESENTING  RELATIVE  ROTATION. 


157 


or  //.  Since;  for  oily  metallic  surfaces,  ft  may  have  a  value 
as  low  as  0.08,  it  follows  that  a  must  not  exceed  4^°.  If  both 
surfaces  of  the  cotter  slope  with  reference  to  the  line  of  action 
of  the  force,  the  total  angle  of  the  sloping  sides  must  not 
exceed  9°. 


FIG.  108. 


FIG.  109. 


109.  Set-screws  (see  §  86,  p.  120)  are  frequently  used  to  pre- 
vent relative  rotation.  They  are  inadvisable  for  heavy  duty. 
Experiments  made  by  Professor  Lanza*  with  f-inch  wrought- 
iron  set-screws,  ten  threads  to  the  inch  and  tightened  with  a 
pull  of  75  Ibs.  at  the  end  of  a  lo-inch  wrench,  gave  the 
following  results: 


TABLE  XII. 


Kind  of  End. 


Holding  Power  at 
Surface  of  Shaft.  - 


Ends  perfectly  flat,.  ^  inch  diameter.  .  .  Average  2064  Ibs. 

Rounded  ends,  radius  %  inch  • "         2912     " 

Rounded  ends,  radius  \  inch "         2573     '< 

Cup-shaped  and  case  hardened "         2470    " 

no.  Shrink  and  Force  Fits. —  Relative  rotation  between 
machine  parts  is  also  prevented  sometimes  by  means  of  shrink 
and  force  fits.  In  the  former  the  shaft  is  made  larger  than 
the  hole  in  the  part  to  be  held  upon  it,  and  the  metal  surround- 
ing the  hole  is  heated,  usually  to  low  redness,  and  because  of 
the  expansion  it  may  be  put  on  the  shaft,  and  on  cooling  it 


*  Trans.  A.  S.  M.  E.,  Vol.  X. 


1 58  MACHINE  DESIGN. 

shrinks  and  "  grips  "  the  shaft.  A  key  is  sometimes  used  in 
addition  to  this.  For  proper  allowances  see  the  accompanying 
table.  The  coefficient  of  linear  expansion  for  each  degree 
Fahrenheit  is  0.0000065  for  wrought  iron  and  steel  and 
0.0000062  for  cast  iron.  Low  redness  corresponds  to  about 
600°  F.  and  therefore  causes  an  expansion  of  the  bore  of  about 
0.004  inch  per  inch  of  diameter. 

Force  fits  are  made  in  the  same  way  except  that  they  are 
put  together  cold,  either  by  driving  together  with  a  heavy  sledge 
or  by  forcing  together  by  hydraulic  pressure.  The  necessary 
allowance,  i.e.,  excess  of  shaft  diameter  over  the  diameter  of 
the  hole,  is  given  in  the  following  table  (XIII),  compiled  by 
Mr.  S.  H.  Moore  * 

Column  i  gives  values  of  "  pressure  factors "  which  are  to 
be  used  in  connection  with  forced  fits  to  determine  the  pressure 
necessary  to  force  the  machinery  steel  shaft  into  the  cast-iron 
hub.  The  formula  to  be  used  is 

Area  of         difference  in   dia.  be-  Ypp 

surface  of  fit     tween    plug  and    bore 

Pressure  in  tons  = 

2 

In  using  this  table  for  forced  fits  the  surfaces  should  be  as 
smooth  as  possible;  if  they  are  ground  it  is  best.  They  should 
be  well  lubricated.  Mr.  Moore  recommends  linseed-oil  for  this 
purpose. 

Compared  with  average  practice  the  forcing-fit  allowances  of 
this  table  are  too  large.  Satisfactory  results  will  be  obtained 
by  using  just  one  half  these  tabular  values. 

In  general,  pressure  fits  are  not  employed  on  diameters  ex- 
ceeding 10  ins.,  shrinkage  fits  being  used  for  large  work. 

*  Trans.  A.  S.  M.  E.,  Vol.  XXIV. 


MEANS  FOR  PREVENTING  RELATIVE  ROTATION.  159 

TABLE  XIII.— DATA  RELATIVE  TO  FITS  AND  FITTING. 


Pressure 
Factors. 

Nominal 
Diameter 
of  Fit. 

Forcing  Fit, 
Allowances. 

Shrinking  Fit, 
Allowances. 

Driving  Fit, 
Allowances. 

Running  Fit, 
Allowances. 

.0006 

.00058 

.0008 

.00065 

.0009 

.00073 

391 

I 

.0025 

.0016 

.0010 

.0008 

319 

ii 

•0035 

.0021 

.0013 

.0010 

240 

2 

.0045 

.0026 

.0015 

.0011 

156 

3 

.0065 

.0037 

.0020 

.0014 

"5 

4 

.0085 

.0048 

.0025 

.0018 

9* 

5 

.0105 

.0058 

.0030 

.0021 

75 

6 

.0125 

.0069 

•0035 

.0024 

64 

7 

.0145 

.0079 

.OO4O 

.0027 

51-5 

8 
9 

.0165 
.0185 

.0090 
.0101 

.0045 
0050 

.0030 
•0033 

43 

10 

.0205 

.0111 

•0055 

.0036 

39 

ii 

.0225 

.0122 

.0060 

.0039 

36 

12 

.0245 

•0133 

.0065 

.0043 

30-4 

14 

.0285 

.0154 

.0075 

.0049 

26.4 

16 

•0325 

•0175 

.0085 

•0055 

23-3 

18 

•0365 

.0196 

•0095 

.Oo6l 

20.8 

20 

.0405 

.0218 

.0068 

18.8 

22 

•0445 

.0239 

.0074 

17.2 

24 

.0485 

.0200 



.0080 

T5-i 

27 

•0545 

.0292 

.0090 

13-5 

3° 

.0605 

.0324 

.0099 

38 

.040  ^ 

10 

8  Is 

44 

+  8 

.047  +  § 

r  R 

+  8 

5° 
56 

£  § 

•053  q  8 

.060  H» 

a  ~ 

Q  2 

•43 

i 

[r 

62 

ii 

.066     ,| 

i 

£ 

66 

"< 

.070    ^ 

^ 

t 

The  alignment  should  be  absolutely  accurate  in  starting. 
To  secure  this  some  engineers  resort  to  the  use  of  two  diameters 
— each  half  the  length  of  the  fit — differing  by  but  a  few  thou- 
sandths of  an  inch. 

Experience  shows  that,  for  same  allowances,  shrink  fits  hold 
more  firmly  than  pressure  fits.* 

in.  Stress  in  Hub. — In  regard  to  the  tension  in  the  hub  due 
to  shrinkage  or  forced  fits,  completely  satisfactory  data  are 
lacking.  A  close  approximation  to  the  probable  tension  in  the 
inner  layer  of  hub — the  one  next  the  shaft — may  be  made 

*  For  valuable  data  on  fits  and  fittings,  see  Am.  Much.,  Mar.  7,  1907. 


160  MACHINE  DESIGN. 

by  considering  the  hub  as  a  thick  cylinder  under  internal  pres- 
sure.    Then  * 


'  '  ! 

/i  =unit  stress  in  pounds  per  square  inch  in  inner!  layer; 
pi  =  internal  pressure  in  pounds  per  square  inch; 
r\  =  internal  radius  of  hub  in  inches; 
TZ  =  external  radius  of  hub  in  inches. 

To  determine  the  probable  value  of  pi  we  may  proceed  as 
follows:  Knowing  the  maximum  forcing  pressure  in  tons, 

/area  of  surface  of  fit  X  A  XPF\ 
\~~  ~T~  /' 

this  may  be  presumed  to  be  the  resistance  offered  by  friction 
between  the  surfaces.  Assuming  the  coefficient  of  friction 
=  0.10,  the  total  pressure  between  the  surfaces  is  ten  times  as 
great  as  the  forcing  pressure.  Reducing  this  total  pressure  to 
pounds,  and  dividing  by  the  area  of  surface  of  fit,  gives  the 
pressure  per  square  inch,  or  the  value  of  p{. 

.'.  pi  =  10,000  A  XPF. 

112.  Problem.  —  A  hub  y|  inches  in  diameter  and  8  inches 
long  is  to  be  forced  on  a  5-inch  shaft.  What  allowance  should 
be  made  in  .difference  of  diameter  of  bore  and  shaft?  What 
will  be  the  necessary  forcing  pressure  ?  What  will  be  the  tensile 
stress  in  the  inner  fibers  of  the  hub? 

From  the  table  the  forcing-fit  allowance  for  a  5  -inch  nominal 
diameter  is  0.0105"  =  A. 

*  Swing's  "Strength  of  Materials,"  p.  210. 


MEANS  FOR  PREVENTING  RELATIVE  ROTATION.  161 

The  forcing  pressure  in  tons 

(5X7TX8)  X  (0.0105)  Xoi 

—  —  —  =  59-I3> 

since  the  table  gives  us  PF=gi  for  this  case. 
The  internal  pressure 

^•  =  10,000X0.0105X91  =9555  Ibs.  per  square  inch, 
and 


This  is  a  dangerous  value  for  cast  iron  and  points  to  a  greater 
outside  diameter  of  hub  or  a  smaller  fit  allowance.  The  rough 
rule  of  practice  is  to  make  he  outside  diameter  of  hub  equal 
twice  the  diameter  of  the  shaft  for  cast-iron  hubs  This 
would  reduce  /i  to  17,500  lbs.,  about.  Or,  using  the  original 
hub  but  only  one  half  the  tabular  allowance,  /i  becomes 
12,405  lbs. 


CHAPTER  X. 


SLIDING   SURFACES. 

113.  General  Discussion. — So  much  of  the  accuracy  of 
motion  of  machines  depends  on  the  sliding  surfaces  that  their 
design  deserves  the  most  careful  attention.  The  perfection  of 
the  cross-sectional  outline  of  the  cylindrical  or  conical  forms 
produced  in  the  lathe  depends  on  the  perfection  of  form  of  the 
spindle.  But  the  perfection  of  the  outlines  of  a  section  through 
the  axis  depends  on  the  accuracy  of  the  sliding  surfaces.  All 
of  the  surfaces  produced  by  planers,  and  most  of  those  pro- 
duced by  milling-machines,  are  dependent  for  accuracy  on  the 
sliding  surfaces  in  the  machine. 


FIG.  no. 

114.  Proportions  Dictated  by  Conditions  of  Wear. — Suppose 
that  the  short  block  A,  Fig.  no,  is  the  slider  of  a  slider-crank 
chain,  and  that  it  slides  on  a  relatively  long  guide  D.  The 
direction  of  rotation  of  the  crank  a  is  as  indicated  by  the  arrow. 
B  and  C  are  the  extreme  positions  of  the  slider.  The  pressure 
between  the  slider  and  the  guide  is  greatest  at  the  mid-position, 
A ;  and  at  the  extreme  positions,  B  and  C,  it  is  only  the  pressure 
due  to  the  weight  of  the  slider.  Also  the  velocity  is  a  maximum 

162 


SLIDING  SURFACES.  l63 

when  the  slider  is  in  its  mid -position,  and  decreases  towards  the 
ends,  becoming  zero  when  the  crank  a  is  on  its  center.  The 
work  of  friction  is  therefore  greatest  at  the  middle,  and  is  very 
small  near  the  ends.  Therefore  the  wear  would  be  the  greatest 
at  the  middle,  and  the  guide  would  wear  concave.  If  now  the 
accuracy  of  a  machine's  working  depends  on  the  perfection 
of  A's  rectilinear  motion,  that  accuracy  will  be  destroyed  as  the 
guide  I?  wears.  Suppose  a  gib,  EFG,  to  be  attached  to  A,  Fig. 
in,  and  to  engage  with  D,  as  shown,  to  prevent  vertical  loose- 
ness between  A  and  D.  If  this  gib  be  taken  up  to  compensate 
for  wear  after  it  has  occurred,  it  will  be  loose  in  the  middle 
position  when  it  is  tight  at  the  ends,  because  of  the  unequal 
wear.  Suppose  that  A  and  D  are  made  of  equal  length,  as  in 
Fig.  112.  Then  when  A  is  in  the  mid-position  corresponding 


to  maximum  pressure,  velocity,  and  wear,  it  is  in  contact  with 
D  throughout  its  entire  surface,  and  the  wear  is  therefore  the 
same  in  all  parts  of  the  surface.  The  slider  retains  its  accuracy 
of  rectilinear  motion  regardless  of  the  amount  of  wear;  the 
gib  may  be  set  up,  and  will  be  equally  tight  in  all  positions. 

f  A  I  I  B   I  e 


FIG.  113. 

If  A  and  B,  Fig.  113,  are  the  extreme  positions  of  a  slider, 
D  being  the  guide,  a  shoulder  would  be  finally  worn  at  C.  It 
would  be  better  to  cut  away  the  material  of  the  guide,  as  shown 


1 64  MACHINE  DESIGN. 

by  the  dotted  line.  Slides  should  always  "wipe  over"  the  ends 
of  the  guide  when  it  is  possible.  Sometimes  it  is  necessary 
to  vary  the  length  of  stroke  of  a  slider,  and  also  to  change  its 
position  relatively  to  the  guide.  Examples:  "Cutter-bars" 
•of  slotting-  and  shaping-machines.  In  some  of  these  positions 
there  will  be  a  tendency,  therefore,  to  wear  shoulders  in  the 
Iguide  and  also  in  the  cutter-bar  itself.  This  difficulty  is  over- 
come if  the  slide  and  guide  are  made  of  equal  length,  and  the 
design  is  such  that  when  it  is  necessary  to  change  the  position 
of  the  cutter-bar  that  is  attached  to  the  slide,  the  position  of 
the  guide  may  be  also  changed  so  that  the  relative  position  of 
slide  and  guide  remains  the  same.  The  slider  surface  will  then 
just  completely  cover  the  surface  of  the  guide  in  the  mid-position, 
and  the  slider  will  wipe  over  each  end  of  the  guide  whatever 
the  length  of  the  stroke. 

In  many  cases  it  is  impossible  to  make  the  slider  and  guide 
of  equal  length.  Thus  a  lathe-carriage  cannot  be  as  long  as  the 
bed,  a  planer-table  cannot  be  as  long  as  the  planer-bed,  nor  a 
planer-saddle  as  long  as  the  cross-head.  When  these  condi- 
tions exist  especial  care  should  be  given  to  the  following: 

I.  The  bearing  surface  should  be  made  so  large  in   pro- 
portion to  the  pressure  to  be  sustained  that  the  maintenance 
of  lubrication  shall  be  insured  under  all  conditions. 

II.  The  parts  which  carry  the  wearing  surfaces  should  be 
made  so  rigid  that  there  shall  be  no  possibility  of  the  localiza- 
tion of  pressure  from  yielding. 

115.  Form  of  Guides. — As  to  form,  guides  may  be  divided 
Into  two  classes:  angular  guides  and  flat  guides.  Fig.  114,  a, 
shows  an  angular  guide,  the  pressure  being  applied  as  shown. 
The  advantage  of  this  form  is,  that  as  the  rubbing  surfaces 
wear,  the  slide  follows  down  and  takes  up  both  the  vertical 
and  lateral  wear.  The  objection  to  this  form  is  that  the  pres- 
sure is  not  applied  at  right  angles  to  the  wearing  surfaces,  as 


SLIDING  SURFACES. 


165 


it  is  in  the  flat  guide  shown  in  b.     But  in  &,  a  gib  must  be  pro- 
vided to  take  up  the  lateral  wear.     The  gib  is  either  a  wedge 


or  a  strip  with  parallel  sides  backed  up  by  screws.  Guides 
of  these  forms  are  used  for  planer-tables.  The  weight  of  the 
table  itself  holds  the  surfaces  in  contact,  and  if  the  table  is  light 
the  tendency  of  a  heavy  side  cut  would  be  to  force  the  table 
up  one  of  the  angular  surfaces  away  from  the  other.  If  the 
table  is  very  heavy,  however,  there  is  little  danger  of  this,  and 
hence  the  angular  guides  of  large  planers  are  much  flatter 
than  those  of  smaller  ones.  In  some  cases  one  of  the  guides 
of  a  planer-table  is  angular  and  the  other  is  flat.  The  side 
bearings  of  the  flat  guide  may  then  be  omitted,  as  the  lateral 
wear  is  taken  up  by  the  angular  guide.  This  arrangement  is 
undoubtedly  good  if  both  guides  wear  down  equally  fast. 


FIG.  115. 

Fig.  115  shows  three  forms  of  sliding  surfaces  such  as  are 
used  for  the  cross-slide  of  lathes,  the  vertical  slide  of  shapers, 
the  table  slide  of  milling-machines,  etc.  A  is  a  taper  gib  that 
is  forced  in  by  a  screw  at  D  to  take  up  wear.  When  it  is  neces- 
sary to  take  up  wear  at  B,  the  screw  may  be  loosened  and  a 
shim  or  liner  may  be  inserted  between  the  surfaces  at  a.  C  is 
a  thin  gib  and  the  wear  is  taken  up  by  means  of  several  screws 


166  MACHINE  DESIGN. 

like  the  one  shown.  This  form  is  not  so  satisfactory  as  the 
wedge  gib,  as  the  bearing  is  chiefly  under  the  points  of  the 
screws,  the  gib  being  thin  and  yielding,  whereas  in  the  wedge 
there  is  complete  contact  between  the  metallic  surfaces. 

The  sliding  surfaces  thus  far  considered  have  to  be  designed 
so  that  there  will  be  no  lost  motion  while  they  are  moving, 
because  they  are  required  to  move  while  the  machine  is  'in 
operation.  The  gibs  have  to  be  carefully  designed  and  accu- 
rately set  so  that  the  moving  part  shall  be  just  "  tight  and 
loose  ";  i.e.,  so  that  it  shall  be  free  to  move,  without  lost  motion 
to  interfere  with  the  accurate  action  of  the  machine.  There 
is,  however,  another  class  of  sliding  parts,  like  the  sliding- 
head  of  a  drill-press,  or  the  tailstock  of  a  lathe,  that  are  never 
required  to  move  while  the  machine  is  in  operation.  It  is  only 
required  that  they  shall  be  capable  of  being  fastened  accu- 
rately in  a  required  position,  their  movement  being  simply  to 
readjust  them  to  other  conditions  of  work  while  the  machine 
is  at  rest.  No  gib  is  necessary  and  no  accuracy  of  MOTION 
is  required.  It  is  simply  necessary  to  insure  .that  their  posi- 
ion  is  accurate  when  they  are  clamped  for  the  special  work 
to  be  done. 

116.  Lubrication. — The  question  of  strength  rarely  enters 
into  the  determination  of  the  dimensions  of  sliding  surfaces; 
these  are  determined  rather  by  considerations  of  minimizing 
wear  and  maintaining  lubrication.  As  long  as  a  film  of  oil 
separates  the  surfaces,  wear  is  reduced  to  a  minimum.  The 
allowable  pressure  between  the  surfaces  without  destruction 
of  the  film  of  lubricant  varies  with  several  conditions.  To 
make  this  clear,  suppose  a  drop  of  oil  to  be  put  into  the  middle 
of  an  accurately  finished  surface  plate  (i.e.,  as  close  an  approxi- 
mation to  a  plane  surface  as  can  be  produced) ;  suppose  another 
exactly  similar  plate  to  be  placed  upon  it  for  an  instant;  the 
oil-drop  will  be  spread  out  because  of  the  force  due  to  the  weight 


SLIDING  SURFACES.  167 

of  the  upper  plate.  Had  the  plate  been  heavier  it  would  have 
been  spread  out  more.  If  the  plate  were  allowed  to  remain 
a  longer  time,  the  oil  would  be  still  further  spread  out,  and 
if  its  weight  and  the  time  were  sufficient,  the  oil  would  finally 
be  squeezed  entirely  out  from  between  the  plates,  and  the 
metal  surfaces  would  come  into  contact.  The  squeezing  out 
of  the  oil  is,  therefore,  a  function  of  the  time  as  well  as  of 
pressure. 

If  the  surfaces  under  pressure  move  over  each  other,  the 
removal  of  the  oil  is  facilitated.  The  greater  the  velocity  of 
movement  the  more  rapidly  will  the  oil  be  removed,  and  there- 
fore the  squeezing  out  of  the  oil  is  also  a  function  of  the  velocity 
oj  the  rubbing  surfaces. 

117.  Allowable  Bearing  Pressure. — Flat  surfaces  in  machines 
are  particularly  difficult  to  make  perfectly  true  in  the  first  place, 
and  to  keep  true  in  the  course  of  operation  of  the  machines. 
If  they  are  distorted  ever  so  slightly  the  pressure  between  the 
.surfaces  becomes  concentrated  at  one  small  area,  and  the  actual 
pressure  per  square  inch  is  vastly  in  excess  of  the  nominal 
pressure. 

In  consequence  of  this  and  the  differences  in  original  truth 
and  finish  of  the  surfaces,  there  is  no  matter  in  machine  design 
in  which  practice  varies  more  than  in  the  nominal  pressure 
allowed  per  square  inch  of  bearing  area  of  flat  sliding  surfaces. 

Unwin*  gives  the  following: 

TABLE  XIV. 

Slipper  slide-blocks,  marine  engines too  Ibs.  per  square  inch 

Stationary -engine  slide-blocks 2510125     "      "        " 

Stationary-engine  slide-blocks  usually.  ...    30  to  60     "      : '        " 

Professor  Barr  f  found  American  practice  to  vary  as  follows : 

*  "Machine  Design,"  i4th  ed.,  Vol.  I,  p.  198. 
t  Trans.  A.  S.  M.  E.,  Vol.  XVIII,  p.  753. 


1 68  MACHINE  DESIGN. 

Cross-head  shoes  of  high-speed  engines: 

Minimum  pressure  per  square  inch 10.5  Ibs. 

Maximum  pressure  per  square  inch 38        " 

Mean  pressure  per  square  inch 27        " 

Cross-head  shoes  of  low-speed  engines: 

Minimum  pressure  per  square  inch 29  Ibs. 

Maximum  pressure  per  square  inch 58    " 

Mean  pressure  per  square  inch 40    ' ' 

In  all  cases  the  mean  sliding  velocity  was  probably  in  the 
neighborhood  of  600  feet  per  minute  with  a  maximum  velocity 
at  the  middle  of  the  stroke  of  about  950  feet  per  minute.  In 
"  low-speed  "  engines  the  maximum  velocity  is  only  reached 
about  one  third  as  many  times  per  minute  as  in  "  high-speed  " 
engines,  although  they  may  have  the  same  mean  velocity,  and 
it  is  therefore  proper  to  allow  a  higher  unit  value  of  pressure 
for  the  former  than  for  the  latter.  For  well-made  surfaces 
the  maximum  values  given  by  Professor  Barr  may  be  safely 
used. 

For  lower  mean  speeds  than  600  feet  per  minute  they  may 
be  increased,  and  for  higher  speeds  decreased,  according  to  some 
law  such  as 

^7  =  36,000, 

in  which  formula  p=  pressure  per  square  inch,  and  V  =  velocity 
of  rubbing  in  feet  per  minute. 

118.  Maintenance  of  Lubrication. — Regarding  the  materials 
to  be  used,  brass,  bronze,  or  babbitt  metal  will  run  well  with 
iron  or  steel. 

To  maintain  lubrication  a  constant  flow  of  oil  from  a  cup  is 
desirable.  The  moving  surface  should,  if  possible,  have  chan- 
nels cut  in  its  face  to  conduct  the  oil  from  the  central  oil-hole  to 
all  parts  of  the  surface,  as  shown  in  Fig.  116.  The  oil  should 
be  forced  in  where  the  pressures  are  heavy. 

Oil-pads  may  be  used  as  shown  in  Fig.  117.    The  shaded 


SLIDING  SURFACES. 


169 


areas  represent  porous  pads  whose  lower  surfaces  just  touch  the 
surface  to  be  lubricated,  and  which  are  kept  soaked  with  oil. 


FIG.  117. 


For  oiling  the  ways  of   planer-tables  it  is  customary  to  use 
rollers  placed  in  oil-filled  pockets  in  the  guides.     The  top  of 


FIG.  118. 


the  roller  is  held  against  the  surface  of  the  way  by  means  of 
springs.     (See  Fig.  118.) 


CHAPTER  XI. 

AXLES,  SHAFTS,  AND  SPINDLES. 

up.  By  Axles,  Shafts,  or  Spindles  we  denote  those  rotating 
or  oscillating  members  of  machines  whose  motion  is  constrained 
by  turning  pairs.  Axle  is  the  name  given  to  such  a  member 
when  it  is  subjected  to  a  load  which  produces  a  bending  moment, 
and  the  only  torsional  stress  is  that  due  to  friction. 

When  rotating  members  are  subjected  chiefly  to  torsional 
stress,  or  combined  torsion  and  bending,  they  are  called  shafts 
or  spindles  The  former  term  is  used  where  the  part  has  as 
its  function  the  transmission  of  energy  of  rotation  from  one 
point  to  another.  Examples  are  line-shafts  and  crank-shafts. 

The  term  spindle,  on  the  other  hand,  is  restricted  to  those 
rotating  members  which  are  directly  connected  with  the  tool  or 
work  and  give  it  an  accurate  rotative  motion.  They  generally 
form  the  main  axis  of  the  machine.  Examples  are  the  lathe  and 
drill-spindles. 

120.  Axle  Design. — The  question  of  axle  design  will  be  taken 


FIG.  119. 

up  first,  and  the  torsional  moment  due  to  friction  will  be  neg- 
lected. 

A  typical  case  is  shown  in  Fig.  119.     Here  the  two  ends  are 

170 


AXLES,  SHAFTS,  AND  SPINDLES. 


171 


purposely  not  symmetrical.  Given  the  loads  P  and  Q,  solution 
is  first  made  for  the  reactions  R  and  S  by  the  ordinary  methods 
of  mechanics,  graphical  or  analytical. 

The  graphical  method  is  best  since  it  gives  the  moments  at 
all  sections.  Lay  off  the  line  M-N,  Fig.  120,  whose  length 
equals  I'  +1  +  1",  the  distance  between  the  points  of  application 
of  R  and  S.  Denote  the  points  of  application  of  R,  P,  Q,  and 
S  by  a,  b,  c,  and  d,  respectively.  At  b  erect  a  perpendicular  and 


FIG.  120. 


ky  off  on  it  a  vector  representing  the  value  of  P  in  pounds.  At 
c  erect  a  perpendicular  and  lay  off  on  it  a  vector  representing 
Q  on  -the  same  scale.  At  d  drop  a  perpendicular  and  lay  off 
de  equal  to  vector  Q,  and  ef  equal  to  vector  P.  Select  any 
point  O  as  pole,  and  draw  Od,  Oe,  and  Of.  Denote  by  g  the  point 
where  O  d  intersects  a  perpendicular  dropped  from  c,  and  draw 
from  g  a  parallel  to  Oe  until  it  intersects  a  perpendicular  dropped 
from  b  at  h.  From  h  draw  a  parallel  to  Of  until  it  intersects  a 
perpendicular  dropped  from  a  at  j.  Draw  jd,  and  parallel  to  jd 
draw  a  line  through  O.  This  line  cuts  the  perpendicular  dropped 
from  d  at  the  point  k.  Then  vector  }k=R,  and  kd  =  S,  on 
the  same  scale  as  was  originally  used  for  P  and  Q.  Values 
of  R  and  S  in  pounds  are  therefore  determined.  The  shaded 
area  dghjd  is  the  moment  diagram.  The  vertical  ordinates 


172  MACHINE  DESIGN. 

included  between  its  bounding  lines  are  proportional  to  the 
moments  at  the  corresponding  points. 

The  scale  of  the  moment  diagram  can  be  readily  determined 
by  solving  for  the  actual  moment  for  one  section.  Select  the  sec- 
tion at  b.  The  moment  Mb  is  represented  by  mh  and  has  a 
value  =Rl',  R  being  expressed  in  pounds  and  V  in  inches;  the 
value  of  the  moment  in  inch-pounds  can  be  determined  at  any 
point,  since  the  scale  used  is  mh  inches  equals  Rl'  inch-pounds. 

For  a  circular  section  we  have  the  elastic  moment 

if.-M.fe!!!. 

c         4 

Mb  is  the  bending  moment  in  inch-pounds; 
ft  is  the  unit  stress  in  outer  fiber  in  pounds  per  square  inch; 
/  is  the  plane  moment  of  inertia  of  the  section  in  biquad- 
ratic inches; 
c  is  the  distance  from  neutral  axis  to  outermost  fiber  in 

inches. 

Equating  this  to  the  various  selected  values  of  Mb  and  solv- 
ing for  r  gives  the  radius  of  the  axle  at  any  point. 

In  designing  axles,  great  care  must  be  taken  that  all  forces 
acting  are  being  considered,  and  that  the  maximum  value  of 
each  is  selected.* 

Thus  it  has  been  found  that  the  force  due  to  vertical  oscilla- 
tion caused  by  jar  in  running  is  about  40  per  cent  of  the  static 
load  for  car-axles.  The  axles  would  therefore  have  to  be  de- 
signed for  a  load  1.4  times  the  static  load.  In  addition  to  this 
there  is  in  car-axles  a  bending  moment  due  to  curves,  switches, 
and  wind-pressures.  This  may  amount  as  a  maximum  to  the 
equivalent  of  a  horizontal  force  H,  equal  to  40  per  cent  of  the 
static  load,  applied  at  a  height  of  6  feet  from  the  rail. 

*  See  further  Proc.  Master  Car-Builders'  Assn.,  1896;  Report  of  Committee 
on  Axles,  etc.  Also  Strength  of  Railway-Car  Axles,  Trans.  A.  S.  M.  E.,  1895. 
Reuleaux,  "The  Constructor,"  trans,  by  H.  H.  Supplee,  Philadelph  a,  1893; 
Railway  Machin  -ry,  Mar.  1907. 


AXLES,  SHAFTS,  AND  SPINDLES.  173 

When  such  a  careful  analysis  of  the  forces  has  been  made, 
It  may  be  taken  for  good  material,  equal  to  one  fourth  of  the 
ultimate  strength,  this  being  a  safe  value  for  cases  like  this, 
where  the  fibers  are  subjected  to  alternate  tension  and  com- 
pression as  determined  by  Wohler  and  others. 

Had  the  static  load  alone  been  considered  in  the  calculations, 
It  should  not  have  been  taken  greater  than  one  tenth  of  the 
ultimate  strength. 

121.  Shafting  Subject  to  Simple  Torsion.  —  If  a  short  shaft 
is  subjected  to  simple  torsion,  its  diameter  may  be  deter- 
mined very  readily  by  the  simple  formula  for  torsional 
moment, 


Here  Mt=PR  is  the  torsional  moment,  P  being  the  force  tending 
to  twist  the  piece  in  pounds  and  R  being  the  lever-arm  of  P 
about  the  axis  of  the  piece  in  inches. 

/  is  the  polar  moment  of  inertia  of  the  cross-section  of   the 

member  in  biquadratic  inches; 
c  is  the  distance  from  the  neutral  axis  to  the  outermost  fiber 

in  inches; 

fa  is  the  allowable  unit  stress  in  pounds  per  square  inch. 
For  a  solid  circular  section 


and  M,=!?f, 


which  can  be  solved  readily  for  r,  the  radius  of  the  shaft. 
For  a  hollow  circular  (i.e.,  ring-shaped)  section, 


MACHINE  DESIGN. 

/_g(fl4-ra4) 
C  2f! 


and 


Here  PI  is  the  radius  of  the  outside  of  the  shaft  and  r%  is  the 
radius  of  the  bore,  both  in  inches. 

The  way  to  solve  this  is  to  take  7*2  as  a  decimal  part  of  r\. 
Thus,  let  r2  =  br\.  It  then  becomes  an  easy  matter  to  solve 
for  r\. 

122.  Shafting  Subject  to  Combined  Torsion  and  Bending. — 
In  most  cases  shafts  are  subjected  to  combined  torsion  and 
bending.  Consider  the  crank-shaft  shown  in  Fig.  121  in  side 
and  end  view. 


B  is  the  center  of  the  bearing,  C  is  the  center  of  the  crank- 
pin.  At  B  we  have  the  shaft  subjected  to  a  bending  moment, 
Mi  =  Pl,  and  also  to  a  twisting  moment,  Mt=PR- 

Let  Meb  represent  the  bending  moment  which  would  produce 
the  same  stress  in  the  outer  fiber  as  Mb  and  Mt  combined.  It 
will  be  called  the  equivalent  bending  moment.  Then  it  has 
been  found  *  that 


*  This  corresponds  to  most  recent  investigations.  See  Bach,  "Elasticitat  uncl 
Festigkeit."  For  simplicity's  sake  Bach's  coefficient  «0  has  been  dropped,  as 
it  modifies  the  result  very  slightly  for  wrought  iron  and  steel. 


AXLES,  SHAFTS,  AND  SPINDLES.  175 


Also,  Meb 


For  a  circular  section  (2)  becomes 
Meb=— 
and  substitution  in  (i)  gives 

which  can  readily  be  solved  for  r,  )  being  given  a  value  equal 
to  the  maximum  allowable  unit  tensile  stress  for  the  material 
and  conditions. 

For  a  hollow  circular  section 


To  solve  this  express  rz  as  a  decimal  part  of  r\.  Substitute  and 
solve  for  r\. 

If  there  are  several  forces  acting,  as  there  are  apt  to  be,  the 
method  is  as  follows:  First,  find  Mb  due  to  all  the  bending 
forces  combined.  Second,  find  Mt  due  to  all  the  twisting  forces 
combined.  Third,  use  these  values  of  Mb  and  Mt  in  equations 
(i),  (3),  and  (4).  Among  the  forces  acting  we  must  not  fail 
to  include  the  weight  of  the  shaft  and  attached  parts. 

123.  Comparison  of  Solid  and  Hollow  Shafts. — It  is  evident 
from  (3)  and  (4)  that  the  dimensions  of  a  solid  shaft  and  a 
hollow  shaft  of  equal  strength  will  have  the  relationship 


1  76  MACHINE   DESIGN. 

If  f2  =  o.6ri,  we  have 


Hence  a  hollow  shaft  whose  internal  bore  is  0.6  of  its  ex- 
ternal diameter,  in  order  to  have  the  same  strength  as  a  solid 
shaft  must  have  its  external  diameter  1.047  times  the  diameter 
of  the  solid  shaft.  The  weight  of  the  hollow  shaft  will  be  70 
per  cent  of  that  of  the  solid  shaft.  It  is  obvious  that  a  con- 
siderable saving  in  weight  may  be  effected  without  appreciable 
increase  in  size  if  the  hollow  section  is  adopted.  By  using  nickel 
steel  in  connection  with  the  hollow  section  we  can  get  com- 
bined maximum  strength  and  lightness.* 

124.  Angular   Distortion.  —  The  angle  by  which  a  shaft  sub- 
jected to  torsion  is  twisted  is  often  an  important  matter.     Let 
this  angle  be  represented  by  $.     Then 

MtLiSo 

*=^r  .....  ...  (5) 

$  =  angle  of  torsion  in  degrees  ; 
Mt  =  twisting  moment  in  inch-pounds; 
L  =  length  of  shaft  in  inches  ; 
J  =  polar  moment  of  inertia  in  biquadratic  inches; 

modulus  of  elasticity 

G  =  modulus  of  torsion  =  -  -,  -  -  . 

2.6 

125.  Combined  Thrust  and  Torsion.  —  When   a  shaft   is  sub- 
jected to  combined  thrust  and  torsion,  the  following  formula 
has  been  developed  by  Prof.  A.  G.  Greenhillrf 

*  See  "Nickel  Steel,"  a  paper  by  D.  H.  Browne  in  Vol.  29  of  the  Trans. 
A.  I  M.  E. 

t  See  Proc.  of  Inst.'of  M.  E.,  1883,  p.  182. 


AXLES,  SHAFTS,  AND  SPINDLES.  ]?7 


/2 


/=the  length  of  shaft  between  bearings  in  inches; 
P  =  the  end  thrust  in  pounds; 
E  =  modulus  of  elasticity; 
7  =  plane  moment  of  inertia  of  the  section  in   biquadratic 

inches  ; 
Mt  =  twisting  moment  in  inch-pounds. 

This  formula  is  strictly  only  applicable  to  vertical  shafts,  as  it 
ignores  the  important  item  of  bending  due  to  the  weight  of  the 
shaft  and  attached  parts. 

126.  Line-shafts.  —  Line-shafts  are  long  shafts  used  to  trans- 
mit power.  They  are  made  of  lengths  coupled  together  and 
supported  by  bearings  at  suitable  intervals.  Pulleys  or  gears 
are  keyed  to  them,  and  should  always  be  placed  as  close  to  the 
supporting  bearings  as  possible. 

Consider  first  a  length  of  such  a  shaft  which  is  subjected  to 
pure  torsion  only,  no  pulleys  being  mounted  on  it. 

Because  of  its  weight,  the  length  of  shaft  between  a  pair  of 
bearings  will  sag  so  that  its  axis  will  not  be  a  straight  line. 

When  a  shaft  revolves  at  a  high  speed,  its  own  inertia  gives 
it  a  tendency  to  instability  independent  of  the  torsion  to  which 
the  shaft  is  subjected.  This  is  due  to  the  action  of  centrifugal 
force. 

The  "sag"  of  the  shaft  causes  the  center  of  mass  to  lie  off 
the  axis  of  rotation.  At  a  certain  speed  the  centrifugal  force  is 
just  sufficient  to  keep  the  shaft  bent.  As  this  critical  speed  is 
passed  the  rapidly  increasing  centrifugal  force  exceeds  the  elastic 
forces  of  the  material  and  the  bending  becomes  large.  The 
shaft  is  then  said  to  "whirl." 

The  relation  existing  between  the  speed,  size  of   shaft,  and 


178  MACHINE  DESIGN. 

distance  between  bearings  which  gives  rise  to  whirling  in  wrought 
iron  or  steel  shafts  is  as  follows : 


L  = 


This  applies  to  an  unloaded  shaft  with  bearings  which  do  not 
fix  the  direction  of  the  ends. 

L=  distance  between  bearings  in  inches; 

r  =  radius  of  shaft  in  inches  ; 

n  =  revolutions  of  shaft  per  second. 
This  becomes  a  matter  of  importance  in  rapidly  rotating  shafts.. 

When  there  are  pulleys  on  the  shaft  the  value  of  L  naturally 
becomes  smaller.* 

If  line-shafts  are  designed  wholly  for  strength,  i.e.,  if  d  is 
determined  by  means  of  equation  (3),  owing  to  their  length, 
there  is  apt  to  be  an  excessive  angular  distortion  $.  It  is 
therefore  customary  to  design  them  for  stiffness  and  check  for 
strength  afterwards. 

$  should  not  exceed  0.075°  Per  f°°t  °f  shaft. 

Combining  this  rule  with  the  formula  (5)  for  angular  dis- 
tortion, 


for  a  round  wrought-iron  or  steel  solid  shaft  we  get 


*  For  a  very  complete  discussion  of  "  Whirling  and  Vibration  of  Shafting," 
see  an  article  by  Professor  Dunkerly  in  the  Phil.  Trans,  of  the  Roy.  Soc.  of 
London,  Vol.  185^,  Part  I,  p.  279. 

t  Counting  on  the  fact  that  the  average  load  is  hss  than  the  maximum,  makers 

|  H  p 
of  steel  shafting  use  values  as  small  as  d=T>.'j.*\  — — 


AXLES,  SHAFTS,  AND  SPINDLES.  1 79 

when  d= diameter  of  shaft  in  inches; 

H.P.  =  horse-power  to  be  transmitted; 
N  =  revolutions  per  minute  of  shaft. 

Having  determined  the  diameter  which  will  give  sufficient 
stiffness  against  torsion  the  allowable  distance  between  sup- 
porting bearings  must  be  calculated.  The  rule  of  practice  is 
to  limit  the  deflection  to  i/ioo  of  an  inch  to  a  foot  of  length. 

Consider  first  a  bare  shaft.     There  are  three  cases : 

ist.  Both  ends  of  the  shaft  are  free  to  take  any  direction. 

2d.  One  end  is  free  and  one  fixed. 

3d.  Both  ends  are  fixed. 

In  each  case 

L  =  length  of  span  in  inches ; 
w  =  weight  of  shaft  per  inch; 
y  =  maximum  deflection; 

-y-  =  average  deflection  per  foot  of  length 

=  1/100  inch; 

L 

.'.  y  = . 

2400 

Each  case  is  that  of  a  uniformly  loaded  beam  with  a  load  =  wL. 
For  case  I,     the  deflection  y = — j=j. 


For  case  II,    the  deflection  y 
For  case  III,  the  deflection  y  •• 


0   UT. 

384^,7 


Since   y=  -  ,    and   for   round   shafting   I=-r-    and    «/== 
2400  64 

.28  —  ,  while  £  =  30,000,000,  it  follows  that 


l8o  MACHINE  DESIGN. 

(Case  I)  L=  75^; 

(Case  II)  L 

(CaseTII)  £ 


When  there  are  loads  due  to  belt  pull,  etc.,  the  deflection 
must  be  determined  in  each  case.  For  ordinary  purposes, 
with  the  average  number  of  pulleys  and  amount  of  belt  pull, 
it  is  safe  to  take  for  loaded  shafts  -£-$  of  the  value  of  L  deter- 
mined for  bare  shafting  for  the  same  conditions.  Case  II 
corresponds  most  closely  to  ordinary  conditions. 


CHAPTER  XII. 

JOURNALS,   BEARINGS,   AND   LUBRICATION.* 

127.  General  Discussion  of  Journals  and  Bearings. — Jour- 
nals and  the  bearings  or  boxes  with  which  they  engage  are 
the  elements  used  to  constrain  motion  of  rotation  or  vibration 
about  axes  in  machines.  Journals  are  usually  cylindrical,  but 
may  be  conical,  or,  in  rare  cases,  spherical.  The  design  of 
journals,  as  far  as  size  is  concerned,  is  dictated  by  one  or 
more  of  the  four  following  considerations. 

(1)  To  provide  for  safety  against  rupture  or  excessive  yield- 
ing under  the  applied  forces. 

(2)  To  provide  for  maintenance  of  form 

(3)  To  provide  against  the  squeezing  out  of  the  lubricant. 

(4)  To  provide  against  overheating. 

To  illustrate  (i),  let  Fig.  122  represent  a  pulley  on  the 
end  of  an  overhanging  shaft  driven  by  a  belt,  ABC.  Rota- 
tion is  as  indicated  by  the  arrow,  and  the  belt  tensions  are  Ti 
and  TV  The  journal,  /,  engages  with  a  box  or  bearing.  D. 
The  following  stresses  are  induced  in  the  journal:  TORSION, 
measured  by  the  torsional  moment  (T\-T^)r.  FLEXURE. 
measured  by  the  bending  moment  (T\\-Tz)a.  This  assumes 
a  rigid  shaft  or  a  self-adjusting  box.  SHEAR,  resulting  from 
the  force  Ti  +  T2.  This  journal  must  therefore  be  so  designed 
that  rupture  or  undue  yielding  shall  not  result  from  these 
stresses. 

To  illustrate  (2),  consider  the  spindle  journals  of  a  grind- 
in  g-lathe.  The  forces  applied  are  very  small,  but  the  FORM 
of  the  journals  must  be  maintained  to  insure  accuracy  in  the 
product  of  the  machine.  A  relatively  large  wearing  surface 

*  See  further  Vol.  27  Trans.  A.  S.  M.  E.,  p.  420-505. 

181 


182  MACHINE  DESIGN. 

is  therefore   necessary,   and   careful   provision  must   be    made 
to  exclude  dust  and  grit. 


FIG.  122. 

To  illustrate  (3),  the  pressure  upon  a  journal  resulting  from 
the  applied  forces  may  be  sufficiently  great  to  squeeze  out  the 
lubricant.  Metallic  contact,  heating,  and  abrasion  of  the  sur- 
faces would  result.  In  what  follows,  the  area  of  a  journal  means 
its  PROJECTED  area;  i.e.,  its  length  multiplied  by  its  diameter. 

128.  Allowable  Bearing  Pressure. — The  allowable  pressure 
per  square  inch  of  area  of  a  journal  varies  with  several  condi- 
tions. The  illustration  of  the  drop  of  oil  between  two  sur- 
face plates,  given  in  the  discussion  of  sliding  surfaces  (p.  166), 
applies  here  also.  The  squeezing  out  of  the  oil  from  between 
the  rubbing  surfaces  of  a  journal  and  its  box  is,  therefore, 
a  function  of  the  TIME  as  well  as  of  pressure.  If  the  sur- 
faces under  pressure  move  over  each  other,  the  removal  of 
the  oil  is  facilitated.  The  greater  the  velocity  of  movement, 
the  more  rapidly  will  the  oil  be  removed,  and  therefore  the 
squeezing  out  of  the  oil  is  also  a  function  of  the  VELOCITY  OF 

JRUBBING  SURFACES. 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  183 

When  a  journal  is  subjected  to  continuous  pressure  in  one 
direction,  as,  for  instance,  in  a  shaft  with  a  constant  belt  pull, 
or  with  a  heavy  fly-wheel  upon  it,  this  pressure  has  sufficient 
time  to  act,  and  is  therefore  effective  for  the  removal  of  oil. 
But  if  the  direction  of  the  pressure  is  periodically  reversed,  as 
in  the  crank-pin  of  a  steam-engine,  the  time  of  action  is  less,  the 
tendency  to  remove  the  oil  is  reduced,  and  the  oil  has  opportu- 
nity to  return  between  the  surfaces.  Hence  a  higher  pres- 
sure per  square  inch  of  journal  would  be  allowable  in  the  second 
-case  than  in  the  first. 

If  the  direction  of  motion  is  also  reversed,  as  in  the  cross- 
head  pin  of  a  steam-engine,  the  oil  has  not  only  an  opportu- 
nity to  return  between  the  surfaces,  but  is  assisted  in  doing 
•so  by  the  reversed  motion.  Therefore  a  still  higher  pressure 
per  square  inch  of  journal  is  allowable.  Practical  experi- 
ence bears  out  these  conclusions. 

The  allowable  pressure  depends  also  upon  the  workmanship 
as  shown  in  the  fit  of  journal  and  box  and  the  condition  of  the 
bearing  surfaces.  The  value  to  be  used  in  each  case  must  be 
decided  by  the  judgment  of  the  designer.  The  following  table 
based  upon  practice  may  be  taken  as  a  guide: 

TABLE  XV. — ALLOWABLE  JOURNAL  PRESSURES. 

Pressure  in  Lbs. 

Kind  of  Bearing.  per  Sq.  In.  of 

Projected  Area. 
Bearings  for  slow  speed  and  intermittent  load,  such  as 

crank-pins  of  shearing-machines 2000-3000* 

Main  journals,  center-crank  high-speed  engines 180—  240 

Main  journals,  side-crank  low-speed  engines 160—  220 

Crank-pins  of  high-speed  engines 250-  600 

Crank-pins  of  low-speed  engines 870-1550 

Cross-head  pins  of  high-speed  engines 910—1675 

Cross-head  pins  of  low-speed  engines 1000-1860 

Car-axle  journals 300-  600 

*  In  Vol.  27.  Trans.  A.S.M.E.  pp.  496-497,  Mr.  Oberlin  Smith  gives  examples 
of  journal  pressures  in  presses  running  as  high  as  20,000  pounds  per  square  inch 
o.i  hardened  steel  toggle  pins;  and  7000  pounds  per  square  iiich,  at  a  surface  speed 
of  140  feet  per  minute,  against  the  cast  iron  pitman  driving  the  ram.  The  journal 
pressure  of  the  main  shaft  of  the  second  press  was  2400  pounds  per  square  inch. 


1 84  MACHINE  DESIGN. 

129.  Heating  of  Journals. — To  illustrate  (4),  even  if  the 
conditions  are  such  that  the  lubricant  is  retained  between  the 
rubbing  surfaces,  heating  may  occur.  There  is  always  a  fric- 
tional  resistance  at  the  surface  of  the  journal;  this  resistance 
may  be  reduced  (a)  by  insuring  accuracy  of  form  and  perfec- 
tion of  surface  in  the  journal  and  its  bearings;  (&)  by  insuring 
that  the  journal  and  its  bearings  are  in  contact,  except  for  the 
film  of  oil,  throughout  their  entire  surface,  by  means  of  rigidity 
of  framing  or  self-adjusting  boxes,  as  the  case  may  demand ; 
(c)  by  selecting  a  suitable  lubricant  to  meet  the  conditions  and 
maintaining  the  supply  to  the  bearing  surfaces.  By  these 
means  the  friction  may  be  reduced  to  a  very  low  value,  but  it 
cannot  be  reduced  to  zero. 

There  must  be  some  frictional  resistance,  and  it  is  always 
converting  mechanical  energy  into  heat.  This  heat  raises  the 
temperature  of  the  journal  and  its  bearing.  If  the  heat  thus 
generated  is  conducted  and  radiated  away  as  fast  as  it  is  gener- 
ated, the  box  remains  at  a  constant  low  temperature.  If,  how- 
ever, the  heat  is  generated  faster  than  it  can  be  disposed  of, 
the  temperature  of  the  box  rises  till  its  capacity  to  radiate  heat 
is  increased  by  the  increased  difference  of  temperature  of  the 
box  and  the  surrounding  air,  so  that  it  is  able  to  dispose  of 
the  heat  as  fast  as  it  is  generated.  This  temperature,  necessary 
to  establish  the  equilibrium  of  heat  generation  and  disposal,, 
may  under  certain  conditions  be  high  enough  to  destroy  the 
lubricant  or  even  to  melt  out  a  babbitt-metal  box-lining.  Sup- 
pose now  that  a  journal  is  running  under  certain  conditions 
of  pressure  and  surface  velocity,  and  that  it  remains  entirely 
cool.  Suppose  next  that,  while  all  other  conditions  are  kept 
exactly  the  same,  the  velocity  is  increased.  All  modern  experi- 
ments on  the  friction  in  journals  show  that  the  coefficient  of 
friction  increases  with  the  increase  of  velocity  of  rubbing  sur- 
face (at  speeds  above  100  feet  per  minute).  Therefore  the  in- 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  185 

crease  in  velocity  would  increase  the  frictional  resistance  at  the 
surface  of  the  journal,  and  the  space  through  which  this  resist- 
ance acts  would  be  greater  in  proportion  to  the  increase  in 
velocity.  The  work  of  the  friction  at  the  surface  of  the  journal 
is  therefore  increased  because  both  the  force  and  the  space 
factors  are  increased.  It  is  this  work  of  friction  which  has 
been  so  increased,  that  produces  the  heat  which  tends  to  raise 
the  temperature  of  the  journal  and  its  box.  The  rate  of  gen- 
eration of  heat  has  therefore  been  increased  by  the  increase  in 
velocity,  but  the  box  has  not  been  changed  in  any  way,  and 
therefore  its  capacity  for  disposing  of  heat  is  the  same  as  it 
was  before,  and  hence  the  tendency  of  the  journal  and  its  bear- 
ing to  heat  is  greater  than  it  was  before  the  increase  in  velocity. 
Some  change  in  the  proportions  of  the  journal  must  be  made 
in  order  to  keep  the  tendency  to  heat  the  same  as  it  was  before 
the  increase  in  velocity.  If  the  diameter  of  the  journal  be 
increased,  the  radiating  surface  of  the  box  will  be  proportion- 
ately increased.  But  the  space  factor -of  the  friction  will  be 
increased  in  the  same  proportion,  and  therefore  it  will  be  appar- 
ent that  this  change  has  not  affected  the  relation  of  the  rate 
of  generation  of  heat  to  the  disposal  of  it.  But  if  the  length 
of  the  journal  be  increased,  the  work  of  friction  is  the  same 
as  before  and  the  radiating  surface  of  the  box  is  increased  and 
the  tendency  of  the  box  to  heat  is  reduced.  If,  therefore,  the 
conditions  are  such  that  the  tendency  to  heat  in  a  journal, 
because  of  the  work  of  the  friction  at  its  surface,  is  the  vital 
point  in  design,  it  will  be  clear  that  the  length  of  the  journal 
is  dictated  by  it,  but  not  the  diameter.  The  reason  why  high- 
speed journals  have  greater  length  in  proportion  to  their  diam- 
eter than  low-speed  journals  will  now  be  apparent. 

The  lost  work  per  minute  due  to  friction  may  be  expressed 
by  P/jLxdN,  in  which  P=mean  total  pressure  on  journal, 
j£  =  coefficient  of  friction,  c?=diameter  of  journal,  and  .ZV=revo- 


i86 


MACHINE  DESIGN. 


lutions  per  minute.  This  energy  is  all  converted  into  heat 
which  should  be  dissipated  through  a  surface  which  is  pro- 
portional to  the  projected  area  of  the  journal,  dl.  It  follows 
that,  other  conditions  remaining  constant,  the  projected  area 
should  be  proportional  to  the  heat  generated,  and  we  may  write 


I 


K 


K  must  be  experimentally  determined  for  a  given  set  of  condi- 
tions. Sufficient  data  are  not  available  to  form  a  general  table 
of  values  of  K. 

130.  Journal  Proportions. — The  proportions  of  engine-jour- 
nals *  may  be  seen  in  the  following  table : 

TABLE  XVI. 


Kind  of  Journal. 


Minimum. 

Maximum. 

Average. 

Main  journal,  center-crank  high-speed  engine..  .  - 
Main  journal,  side-crank  low-speed  engine  

2.0 
1-7 

3-0 
2.1 

2.2 

i-9 

Crank  -pin  high-speed  engine  

I.  Of 

Crank  -pin  low-speed  engine  

i.  if 

Cross-head  pin  high-speed  engine  

1.0 

2.0 

1.25 

Cross-head  pin  low-speed  engine  

1  .0 

i-5 

i-3 

131.  Materials  to  be  Used. — Regarding  the  materials  of 
journals  and  their  boxes  the  following  general  statements  may 
be  made.  It  must  be  borne  in  mind  that  the  terms  babbitt, 
brass,  and  bronze  cover  wide  ranges  of  alloys  of  varying  values. 

Cast  iron,  wrought  iron,  soft  steel,  and  hard  steel  will  all  run 

*  Professor  Barr  on  "Current  Practice   in  Engine  Proportions,"  Trans.  A.  S. 
M.  E..  Vol.  18,  p.  737. 

f  Unpublished  data,  same  investigations. 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  187 

well  at  almost  any  speed  on  babbitt  metal.  The  pressure  per 
square  inch  which  an  ordinary  babbitt  bearing  will  stand  when 
running  cool  (i.e.,  at  very  slow  speed),  before  being  squeezed 
out,  has  been  found  to  be  something  over  2000  Ibs.* 

Cast  iron,  wrought  iron,  soft  steel,  and  hard  steel  will  all 
run  well  on  brass  and  bronze.  Brass  and  bronze  of  ordinary 
compositions  will  carry  5000  Ibs.  per  square  inch  without  suffer- 
ing destruction.  Bronze,  however,  is  much  better  than  brass. 

Cast  iron  will  run  on  cast  iron  where,  owing  to  large  bear- 
ing surfaces,  the  unit  pressure  is  light.  Where  the  pressure  and 
speed  are  high,  as  in  engine-journals,  this  will  not  work.f 

In  the  same  way  steel  will  run  on  cast  iron  even  at  high  speeds 
if  the  pressure  is  light.  It  has  been  found  that  steel  will  not 
run  on  cast  iron  in  engine- journals. J 

Wrought  iron,  soft  steel,  and  hard  steel  will  all  run  on  hard 
steel. 

Steel  under  steel  if  hardened  and  polished  will  run  under  as 
high  a  pressure  as  50,000  Ibs.  per  square  inch. 

132.  Calculation  of  Journals  for  Strength. — Journals  gener- 
ally form  parts  of  axles  on  shafts,  and  the  calculation  of  their 
diameter  for  strength  becomes  part  of  the  calculation  of  thf 
shaft.  The  principles  have  been  developed  at  length  in  thf 
preceding  chapter  and  need  not  be  repeated  here. 

If  the  journal  is  so  held  that  it  may  be  considered  as  sub- 
jected to  pure  shearing  stress,  like  the  crank-pin  of  a  center- 
crank  engine,  then 

t*A=P, 

in  which    P  =  total  maximum  load ; 

A  =  total  area  subjected  to  stress; 

fs  =  safe  shearing  stress  for  the  conditions. 

*  C.  F.  Porter,  Trans.  A.  S.  M.  E.,  Vol.  Ill,  p.  227. 
t  Trans.  A.  S.  M.  E.,  Vol.  VI,  pp.  853-854. 


1  88  MACHINE  DESIGN. 

For  a  journal  subjected  to  a  pure  bending  moment, 

«4          1 

which  becomes  Pl=-  —  for  a  solid  circular  shaft.     Pl  =  bending 

4 
moment,  /  =  safe  unit  stress,  and  r  =  radius  of  shaft.     This  can 

readily  be  solved  for  r. 
If  the  journal  be  hollow, 


„, 


4*1 

ri  being  the  external  and  r2  the  internal  diameter. 

For  combined  bending  and  twisting  such  as  the  main  journal 
of  a  side-crank  engine  is  subjected  to,  the  expression  for  a  solid 
journal  is 

]—• 

For  a  hollow  circular  section 
Mri4-r24) 

Mb  being  the  bending  moment  and  Mt  the  twisting  moment. 

In  general  it  will  be  found  that  journals  proportioned  for 
strength  merely  will  not  have  sufficient  area  to  prevent  heating, 
so  this  item  must  not  be  overlooked. 

133.  Problem.  —  Design  the  main  journal  of    a    side-crank 
low-speed  engine. 
Diameter  of  cylinder  =  16  ins. 
Length  of  stroke          =36  ins. 
Net  forward  pressure  =  100  Ibs.  per  square  inch  of  piston  area. 

Suppose  the  engine  capable  of  carrying  full  pressure  to  half- 
stroke. 


JOURNALS,  BEARINGS,  AND  LUBRICATION. 


189 


The  area  of  piston  =201.06  square  inches. 

.*.  total  net  forward  pressure  =  20,106  Ibs. 

At  point  of  maximum  torsional  effect,  which  corresponds  to 
the  position  of  maximum  velocity  of  piston,  no  energy  is  used 
in  accelerating  reciprocating  parts,  and 

Fpvp=Fcvc-, 

Fp=net  forward  force  on  piston; 
vp  =  velocity  of  piston ; 
Fc=  force  on  crank; 
vc  =  velocity  of  crank. 

Since  vc  is  less  than  vp  for  this  position,  Fe  is  greater  than 
Fp,  since  Fe=-*^-. 

Assuming  a  connecting-rod  length  equal  to  five  and  a  half 
crank  lengths  gives  (Appendix)  Fc  =  20,500  Ibs. 

Since  the  crank  length  is  18  inches,  and  at  this  position  the 
crank  and  connecting-rod  are  nearly  at  a  right  angle  with 
each  other,  there  is  a  twisting  moment  at  the  journal  equal  to 

Mt  =  20,500  X 18  =  369,000  inch-lbs. 

There  is  also  a  bending  moment  equal  to  20,500  X  the  dis- 
tance from  center  of  crank-pin  to  center  of  main  journal.     In 
most  cases  this  distance  must  be 
assumed;  for,  although  the  length 
of  the  crank-pin  and  the  thickness 

•»  of  the  crank  may  be  known,  the 

length  of  the  main  journal  is  un- 
Y  known,  since  this  length  and  the 
journal    diameter    are    the    very 
dimensions  sought.    Assume  then 
that  the  crank-pin  is  6  inches  long, 
the  crank  3  inches   thick,  and  the  middle  of  main   journal  6 


J 


190  MACHINE  DESIGN. 

inches  from  the  inner  face  of  crank  as  shown  in  Fig.  123. 
This  will  give  12  inches  as  the  lever-arm;  .'.  the  bending  moment, 
Mb,  =20,500X12  =  246,000  inch-lbs. 

The  equivalent  bending  moment    to  the   combined  actual 
bending  and  twisting  moments 

=  M*  _ 

=  0.35  X  246,000  +  o.65V/246ooo2  +  3690002 
=  374>375  inch-lbs. 


But 


,     4X374375 


For  a  main  shaft   like   this  /  may  be    taken  =  12,000  Ibs. 
per  square  inch  for  steel. 


.  r=  34X374375 

>  7TXI2OOO 

=  3.41  inches; 

.'.  diameter  of  journal  =  2X3.4i=6.82,  say  7  inches. 

The  length  according  to  practice  would  be  about  twice  this 
diameter,*  or  14  inches.  This  would  give  a  projected  area  of 
98  square  inches  and  a  pressure  of  something  over  200  Ibs. 
per  square  inch  of  bearing  due  to  steam-pressure  alone. 

To  get  the  actual  maximum  pressure,  on  the  journal  it 
would  be  necessary  to  know  the  weight  of  the  shaft,  flywheel, 
and  other  attached  parts,  and  properly  combine  the  pressure 
due  to  these  with  the  pressure  due  to  the  steam. 

The  rough  rule  of  practice  for  Corliss  engines  is  to  make 
the  diameter  of  main  journal  equal  to  one  half  the  diameter 
of  the  cylinder. 

*   See  Table  XVI,  p.  186. 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  191 

134.  Problem.  —  Design  the  crank-pin  for  the  same  engine. 
It  will  be  found  that  the  crank-pin  must  be  designed  with  refer- 
ence to  maintaining  lubrication,  and  that  it  will  have  an  excess 
of  strength. 

Allowing  1  200  Ibs.  per  square  inch  of  area,*  and  noting 
from  the  table  that  the  average  practice  for  this  type  of  engine 
is  to  make  the  length  of  the  pin  =  i.iXthe  diameter,  f  it  fol- 
lows that 


-, 

1200 

but  l=i.  id; 

,_     20500 
.'.  i.id2=  —  -  —  , 
1  200 

and  </  =  4  inches,  nearly; 

.'.  l=i.i  X4  =  4^  inches,  say. 

Checking  this  for  strength,  considering  the  pin  subjected 
to  a  bending  moment  P-,  we  write 


*       4 
P  =  20500  Ibs., 

/    4-5 

-=  —  =  2.25  inches; 

2      2 

r=2  inches, 

/  =  stress  in  outer  fiber; 

.    4X20500X2.25 
•'•  /  =  --  ITT  -  '  =73°°lbs.  per  square  inch; 


which  is,  of  course,  a  perfectly  safe  value  for  wrought  iron  or 
steel. 

*  See  Table  XV,  p.  183.  f  See  Table  XVI,  p.  i8i. 


192  MACHINE  DESIGN. 

135.  Problem. — Design  the  cross-head  pin  for  the  same 
engine.  This  pin  also  should  be  designed  for  maintaining 
lubrication.  Allowing  1400  Ibs.  per  square  inch  as  the  per- 
missible pressure  on  the  journal,*  and  noting  that  the  length 
may  be  taken  as  1.3  times  the  diameter  from  average  practice  f 
gives 

a-2&-, 

1400 

20500 
1400  ' 

and  d  =  3!  inches; 

.*.  1  =  4$  inches. 

Checking  this  for  strength  it  is  evident  that  the  only  way 


FIG.  124. 

this  pin  can  fail  is  by  shearing  on  two  surfaces,  A-B  and  D-E 
(see  Fig.  124). 


20500  .     . 

•*•  /•=g    xx    xx — s^^n^o  Ibs.  per  square  inch. 
2X71X2.84 

This  leaves  so  great  a  margin  of  safety  that  some  manu- 
facturers make  the  cross-head  pin  of  two  parts,  an  inner  pin 
of  soft,  resilient  material,  sufficiently  large  to  resist  the  shear- 
ing stress,  and  an  outer  hard-steel  bushing  which  surrounds 
the  soft  pin,  but  is  not  allowed  to  turn  on  it.  The  nature  of 

*  See  Table,  XV,  p.  183.  t  See  Table  XVI,  p.  186. 


JOURNALS,  BEARINGS,  AND  LUBRICATION. 


193 


the  forces  acting  on  a  cross-head  pin  tend  to  wear  it  to  an  oval 
cross-section.  As  such  wear  takes  place  the  bushing  can  readily 
be  given  a  quarter  turn  and  clamped  in  the  new  position. 
(See  Fig.  124.) 

136.  Maintenance    of  Form. — Journals   whose    maintenance 
of  form  is  of  chief  importance  must  be  designed  from  prece- 
dent,   or    according    to    the   judgment    of  the    designer.     No 
theory   can    lead    to   correct   proportions.     In   fact    these    pro- 
portions are  eventually  determined  by  the  process  of  Machine 
Evolution. 

137.  Thrust- journals. — When    a    rotating-machine    part     is 
subjected  to  pressure  parallel  to  the  axis  of  rotation,  "means 
must  be  provided  for  the  safe  resistance  of  that  pressure.     In 
the  case  of  vertical  shafts  the  pressure  is  due  to  the  weight 
of  the  shaft  and  its  attached  parts,    as  the  shafts  of  turbine 
water-wheels  that   rotate  about   vertical  axes.     In  other  cases 
the  pressure  is  due  to  the  working  force,  as  the  shafts  of  pro- 
peller-wheels, the  spindles  of  a  chucking- lathe,  etc.     The  end- 
thrusts  of  vertical  shafts  are  very  often  resisted  by  the  "squared- 
up  "  end  of  the  shaft.     This  is    inserted  in  a  bronze  or    brass 
"bush,"  which  embraces  it    to   prevent   lateral   motion,   as  in 
Fig.   125.     If  the  pressure  be  too  great,  the  end  of  the  shaft 
may  be  enlarged  so  as  to  increase  the  bearing  surface,  thereby 
reducing    the    pressure    per    square    inch.     This    enlargement 


B 

FIG.  126. 


FIG.  125. 

must  be  within  narrow  limits,  however.     (See  Fig.  126.)    AB  is 
the  axis  of  rotation,  and  ACD  is  the  rotating  part,  its  bear- 


194 


MACHINE  DESIGN. 


ing  being  enlarged  at  CD.  Let  the  conditions  of  wear  be  con- 
sidered. The  velocity  of  rubbing  surface  varies  from  zero 
at  the  axis  to  a  maximum  at  C  and  D.  It  has  been  seen  that 
the  increase  of  the  velocity  of  rubbing  surface  increases  both 
the  force  of  the  friction  and  the  space  through  which  that 
force  acts;  it  therefore  increases  the  work  of  the  friction,  and 
therefore  the  tendency  to  wear.  From  this  it  will  be  seen 
that  the  tendency  to  wear  increases  from  the  center  to  the 
circumference  of  this  "radial  bearing,"  and  that,  after  the 
bearing  has  run  for  a  while,  the  pressure  will  be  localized  near 
the  center,  and  heating  and  abrasion  may  result.  For  this 
reason  where  there  is  severe  stress  to  be  resisted,  the  bearing  is 
usually  divided  up  into  several  parts,  the  result  being  what  is 
known  as  a  "collar  thrust-bearing,"  as  shown  in  Fig.  127. 


FIG.  127. 

By  the  increase  in  the  number  of  collars,  the  bearing  surface 
may  be  increased  without  increasing  the  tendency  to  unequal 
wear.  The  radial  dimension  of  the  bearing  is  kept  as  small 
as  is  consistent  with  the  other  considerations  of  the  design. 
It  is  found  that  the  "  tractrix,"  the  curve  of  constant  tangent, 
gives  the  same  work  of  friction,  and  hence  the  same  tendency 
to  wear  in  the  direction  of  the  axis  of  rotation,  for  all  parts  of 
the  wearing  surface.  (See  "  Church's  Mechanics,"  page  181.) 

This  has  been  very  incorrectly  termed  the  "anti-friction" 
thrust-bearing.  This  is  far  from  being  the  case.  The  friction 
work  for  this  and  all  conical  thrust- bearings  can  be  shown 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  195 

readily  to  be  excessive.     Their  one  advantage  is  that  they  are 
easily  adjustable.     In  general  they  are  to  be  avoided. 

The  pressure  that  is  allowable  per  square  inch  of  projected 
area  of  bearing  surface  varies  in  thrust-bearings  with  several 
conditions,  as  it  does  in  journals  subjected  to  pressure  at  right 
angles  to  the  axis.*  Thus,  in  the  pivots  of  turntables,  swing- 
bridges,  cranes,  and  the  like  the  movement  is  slow  and  never 
continuous,  often  being  reversed;  and  also  the  conditions  are 
such  that  "bath  lubrication"  may  be  used,  and  the  allowable 
unit  pressure  is  very  high — equal  often  to  1500  Ibs.  per  square 
inch,  and  in  some  cases  greatly  exceeding  that  value.  The 
following  table  may  be  used  as  an  approximate  guide  in  the 
designing  of  thrust-bearings.  The  material  of  the  thrust-journal 
is  wrought  iron  or  steel,  and  the  bearing  is  of  bronze  or  brass 
(babbitt  metal  is  seldom  used  for  this  purpose).  Bath  lubrica- 
tion is  used,  i.e.,  the  running  surfaces  are  submerged  constantly 
in  a  bath  of  oil. 

TABLE  XVII.t 

Allowable  Unit 

Pressure, 
Pounds  per 

Mean  Velocity  of  Rubbing  Surface,  Feet  per  Minute.       Square  Inch  of 

Projected  Area 

of  the  Rubbing 

Surface. 

Slow  and  intermittent 1500 

50 200 

50  to  100 loo 

100  to- 150 75 

150  to  200 60 

Above  200 50 

If  the  journal  is  of  cast  iron  and  runs  on  bronze  or  brass, 

*  See  Proc.  Inst.  M.  E.,  1888  and  1891,  for  reports  on  experiments  with  thrust- 
bearings. 

t  Reuleaux,  "  Constructor, "  p.  65,  gives  for  steel  on  bronze, 
Slow  moving  pivots,  d  =  0.035  V^- 
Up  to  150  rev.  per  minute,  d  =  0.05  \/P- 
Above  150  rev.  per  minute,  d  =  0.004  \/Pn. 

d  =  diameter  of  shaft  in  inches,  P  =  total  load  in  pounds,  n  =  revs,  per  minute. 
These  formulas  give  higher  unit  pressures  than  the  table  derived  from  the  experi- 
ments of  the  Inst.  M.E.,  which  are  particularly  applicable  to  collar  thrust  bearings. 


i96 


MACHINE  DESIGN. 


the  values  of  allowable  pressure  given  should  be  divided  by 
two. 

The  most  efficient  forms  of  thrust-bearings  are  those*  em- 
ploying the  principles  shown  in  Fig.  128. 

Between  the  end  of  the  shaft  and  the  bottom  of  the  step  a 
series  of  accurately  finished  disks  are  introduced.  The  disks 
are  alternately  hard  steel  and  bronze,  the  top  one  is  fastened  to 
the  shaft,  the  lower  to  the  step,  and  the  rest  are  free.  As  indi- 
cated, each  disk  has  a  hole  through  the  middle  and  radial  grooves 


FIG.  128. 

to  permit  the  lubricant  to  have  access  between  the  disks.  The 
effect  of  centrifugal  force  when  the  shaft  is  rotating  is  to  force 
the  oil  outward  from  between  the  plates  and  upward.  It  is 
collected  in  the  annular  chamber  a-a  and  flows  from  there  down 
the  drilled  passages  back  to  the  bottom  of  the  bearing.  This 
is  equivalent  to  a  continuous  automatic  pump  action  supplying 
oil  to  the  surfaces. 

This  form  of  bearing  reduces  the  relative  motion  between 
successive  surfaces  to  a  minimum,  thereby  allowing  much  higher 
pressures  to  be  carried. 

A  similar  arrangement  of  loose  disks  can  be  used  to  great 
advantage  on  small  propeller  shafts  and  on  worm  shafts. 

When  bearings  have  to  be  used  where  corrosion  or  electro- 

*  See  Trans.  A.  S.  M.  E.,  Vol.  VI,  p.  852,  and  Proc.  Inst.  M.E.,  1888,  p.  184. 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  IQ7 

lytic  action  is  to  be  feared,  as  in  turbine  work,  glass  and  the 
end  grain  of  very  hard  woods  have  been  used  successfully  as 
bearing  materials. 

138.  Problem. — It  is  required  to  design  the  collar  thrust- 
journal  that  is  to  receive  the  propelling  pressure  from  the  screw 
of  a  small  yacht.  The  necessary  data  are  as  follows:  The 
maximum  power  delivered  to  the  shaft  is  70  H.P.;  pitch  of 
screw  is  4  feet;  slip  of  screw  is  20  per  cent;  shaft  revolves  250 
times  per  minute;  diameter  of  shaft  is  4  inches. 

For  every  revolution  of  the  screw  the  yacht  moves  forward 
a  distance  =  4  feet  less  20  per  cent  =  3. 2  feet,  and  the  speed  of 
the  yacht  in  feet  per  minute  =  2 50X3. 2  =  800. 

70  H. P.  =  70X33,000  =  2,310,000  ft.-lbs.  per  minute. 

This  work  may  be  resolved  into  its  factors  of  force  and  space, 
and  the  propelling  force  is  equal  to  2,310,000-^-800  =  2900  Ibs., 
nearly. 

The  shaft  is  4  inches  in  diameter,  and  the  collars  must  project 
beyond  its  surface.  Estimate  that  the  mean  diameter  of  the 
rubbing  surface  is  4.5  inches,  then  the  mean  velocity  of  rubbing 

surface  would  equal  4.5  X  —  X  250  =  294  feet  per  minute.     The 

allowable  value  of  pressure  per  square  inch  of  journal  surface  for  a 
velocity  above  200  feet  per  minute  is  50  Ibs.  The  necessary  area 
of  the  journal  surface  is  therefore  =  2900  -*•  50  =  58  square  inches. 
It  has  been  seen  that  it  is  desirable  to  keep  the  radial  dimen- 
sion of  the  collar  surface  as  small  as  possible  in  order  to  have 
as  nearly  the  same  velocity  at  all  parts  of  the  rubbing  surface 
as  possible.  The  width  of  collar  in  this  case  will  be  assumed  = 
0.75  inch;  then  the  bearing  surface  in  each  collar 

5.52  X*     42X?r 

—  =  23.7-12.5  =  11.2  sq.  in. 
4  4 

Then  the  number  of  collars  equals  the  total  required  area 
divided  by  the  area  of  each  collar  =58-^  11.2  =  5.18,  say  6. 


198  MACHINE  DESIGN. 

139.  Bearings  and  Boxes. — The  function  of  a  bearing  or  box 
is  to  insure  that  the  journal  with  which  it  engages  shall  have 
an  accurate  motion  of  rotation  or  vibration  about  the  given  axis. 
It  must  therefore  fit  the  journal  without  lost  motion;  must 
afford  means  of  taking  up  the  lost  motion  that  results  neces- 
sarily from  wear  ;  must  resist  the  forces  that  come  upon  it 
through  the  journal,  without  undue  yielding  ;  must  have  the 
wearing  surface  of  such  material  as  will  run  in  contact  with 
the  material  of  the  journal  with  the  least  possible  friction  and 
least  tendency  to  heating  and  abrasion;  and  must  usually 
include  some  device  for  the  maintenance  of  the  lubrication.  The 
selection  of  the  materials  and  the  providing  of  sufficient  strength 
and  stiffness  depends  upon  principles  already  considered,  and 
so  it  remains  to  discuss  the  means  for  the  taking  up  of  necessary 
wear  and  for  providing  lubrication. 

Boxes  are  sometimes  made  solid  rings  or  shells,  the  journal 
being  inserted  endwise.  In  this  case  the  wear  can  only  be 
taken  up  by  making  the. engaging  surfaces  of  the  box  and  journal 
conical,  and  providing  for  endwise  adjustment  either  of  the 
box  itself  or  of  the  part  carrying  the  journal.  Thus,  in  Fig.  129, 
the  collars  for  the  preventing  of  end  motion  while  running  are 
jam-nuts,  and  looseness  between  the  journal  and  box  may  be 
taken  up  by  moving  the  journal  axially  toward  the  left. 


FIG.  129.  FIG.  130. 

By  far  the  greater  number  of  boxes,  however,  are  made  in 
sections  and  the  lost  motion  is  taken  up  by  moving  one  or  more 
sections  toward  the  axis  of  rotation.  The  tendency  to  wear  is 
usually  in  one  direction,  and  it  is  sufficient  to  divide  the  box 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  199 

into  halves.  Thus,  in  Fig.  130,  the  journal  rotates  about  the 
axis  O,  and  all  the  wear  is  due  to  the  pressure  P  acting  in  the 
direction  shown.  The  wear  will  therefore  be  at  the  bottom  of 
the  box.  It  will  suffice  for  the  taking  up  of  wear  to  dress  off 
the  surfaces  at  aa,  and  thus  the  box-cap  may  be  drawn  further 
down  by  the  bolts,  and  the  lost  motion  is  reduced  to  an  admis- 
sible value.  "Liners,"  or  "shims,"  which  are  thin  pieces  of 
sheet  metal,  may  be  inserted  between  the  surfaces  of  division 
of  the  box  at  aa,  and  may  be  removed  successively  for  the  lower- 
ing of  the  box-cap  as  the  wear  renders  it  necessary.  If  the  axis 
of  the  journal  must  be  kept  in  a  constant  position,  the  lower 
half  of  the  box  must  be  capable  of  being  raised. 

Sometimes,  as  in  the  case  of  the  box  "for  the  main  journal  of 
a  steam-engine  shaft,  the  direction  of  wear  is  not  constant. 
Thus,  in  Fig.  131,  A  represents  the  main  shaft  of  an  engine. 
There  is  a  tendency  to  wear  in  the  direction  B,  because  of  the 
weight  of  the  shaft  and  its  attached  parts;  there  is  also  a  ten- 
dency to  wear  because  of  the  pressure  that  comes  through  the 
connecting-rod  and  crank.  The  direction  of  this  pressure  is 
continually  varying,  but  the  average  directions  on  forward  and 


FIG.  131.  FIG.  132. 

return  stroke  may  be  represented  by  C  and  D.  Provision  needs 
to  be  made,  therefore,  for  the  taking  up  of  wear  in  these  two 
directions.  If  the  box  be  divided  on  the  line  EF,  wear  will  be 
taken  up  vertically  and  horizontally  by  reducing  the  liners. 
Usually,  however,  in  the  larger  engines  the  box  is  divided  into 
four  sections,  A,  B,  C,  and  D  (Fig.  132),  and  A  and  C  are 
capable  of  being  moved  toward  the  shaft  by  means  of  screws 


200  MACHINE  DESIGN. 

or  wedges,  while  D  may  be  raised  by  means  of  the  insertion 
of  "  shims." 

The  lost  motion  between  a  journal  and  its  box  is  sometimes 
taken  up  by  making  the  box  as  shown  in  Fig.  133.  The  exter- 
nal surface  of  the  box  is  conical  and  fits  in  a  conical  hole  in 
the  machine  frame.  The  box  is  split  entirely  through  at  Ay 
parallel  to  the  axis,  and  partly  through  at  B  and  C.  The  ends 
of  the  box  are  threaded,  and  the  nuts  E  and  F  are  screwed  on. 
After  the  journal  has  run  long  enough  so  that  there  is  an  unal- 
lowable amount  of  lost  motion,  the  nut  F  is  loosened  and  R 
is  screwed  up,  the  effect  being  to  draw  the  conical  box 
further  into  the  conical  hole  in  the  machine  frame;  the  hole 


FIG.  133. 

through  the  box  is  thereby  closed  up  and  lost  motion  is  reduced. 
After  this  operation  the  hole  cannot  be  truly  cylindrical,  and 
if  the  cylindrical  form  of  the  journal  has  been  maintained,  it 
will  not  have  a  bearing  throughout  its  entire  surface.  This  is 
not  usually  of  very  great  importance,  however,  and  the  form  of 
box  has  the  advantage  that  it  holds  the  axis  of  the  journal  in 
a  constant  position.  As  far  as  is  possible  the  box  should  be 
so  designed  as  to  exclude  all  dust  and  grit  from  the  bearing 
surfaces. 

All  boxes  in  self-contained  machines,  like  engines  or  machine 
tools,  need  to  be  rigidly  supported  to  prevent  the  localization 
of  pressure,  since  the  parts  that  carry  the  journals  are  made  as 
rigid  as  possible.  In  line  shafts  and  other  parts  carrying  journals, 
when  the  length  is  great  in  comparison  to  the  lateral  dimensions, 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  201 

some  yielding  must  necessarily  occur,  and  if  the  bo*es  were 
rigid,  localization  of  pressure  would  result.  Hence  "self- 
adjusting"  boxes  are  used.  A  point  in  the  axis  of  rotation  at 
the  center  of  the  length  of  the  box  is  held  immovable,  but  the 
box  is  free  to  move  in  any  way  about  this  point,  and  thus  adjusts 
itself  to  any  yielding  of  the  shaft.  This  result  is  attained  as 
shown  in  Fig.  134.  O  is  the  center  of  the  motion  of  the  box; 


FIG.  134. 

B  and  A  are  spherical  surfaces  formed  on  the  box,  their  center 
being  at  O.  The  support  for  the  box  contains  internal  spherical 
surfaces  which  engage  with  A  and  B.  Thus  the  point  O  is  always 
held  in  a  constant  position,  but  the  box  itself  is  free  to  move  in 
any  way  about  O  as  a  center.  Therefore  the  box  adjusts 
itself  within  limits  to  any  position  of  the  shaft  and  hence  the 
localization  of  pressure  is  impossible. 

In  thrust-bearings  for  vertical  shafts  the  weight  of  the  shaft 
and  its  attached  parts  serves  to  hold  the  rubbing  surfaces  in 
contact  and  the  lost  motion  is  taken  up  by  the  shaft  following 
down  as  wear  occurs.  In  collar  thrust-bearings  for  horizontal 
shafts  the  design  is  such  that  the  bearing  for  each  collar  is 
separate  and  adjustable.  The  pressure  on  the  different  collars 
may  thus  be  equalized.* 

140.  Lubrication  of  Journals. — The  best  method  of  lubrica- 
tion is  that  in  which  the  rubbing  surfaces  are  constantly  sub- 
merged in  a  bath  of  lubricating  fluid.  This  method  should  be 

*  For  complete  and  varied  details  of  marine  thrust-bearings  see  "  Maw's  Mod- 
ern Practice  in  Marine  Engineering." 


202  MACHINE  DESIGN. 

employed  wherever  possible  if  the  pressure  and  surface  velocity 
are  high.  Unfortunately  it  cannot  be  used  in  the  majority  of 
cases.  It  is  not  necessary  that  the  whole  surface  be  sub- 
merged. If  a  part  of  the  moving  surface  runs  in  the  oil-bath  it 
is  sufficient.*  The  same  result  is  accomplished  by  the  use  of 
chains  and  rings  encircling  the  journals  and  dipping  into  oil- 
pockets,  as  described  later  in  this  section.  The  effect  is  to 
form  a  complete  film  of  oil  enveloping  the  journal.  To  allow 
this  it  is  evident  that  the  bore  of  the  bearing  must  be  slightly 
greater  than  the  diameter  of  the  journal  and  a  series  of  '  'running- 
fit  allowances"  will  be  found  in  Table  XIII,  §  no.f  These 
should  be  increased  with  increase  of  running  speeds. 

The  oil  film  may  be  conceived  to  be  made  up  of  a  series  of 
layers,  the  one  next  the  bearing  surface  remaining  stationary 
with  regard  to  it,  while  the  layer  in  immediate  contact  with  the 
shaft  rotates  with  the  latter.  The  intermediate  layers,  therefore, 
slip  upon  each  other  as  the  shaft  rotates  and  the  friction  becomes 
very  closely  akin  to  "fluid  friction"  with  the  bearing  floating 
upon  the  lubricant,  there  being  no  contact  between  the  metallic 
surfaces.  Fig.  134  A  shows  the  conditions  of  pressure  existing 
in  the  film  in  Tower's  classic  experiments.  It  is  impossible 
to  introduce  oil  satisfactorily  at  the  points  where  the  film  is 
under  pressure;  it  should  be  introduced  and  distributed  where 
the  pressure  is  least.  Under  the  action  of  the  load  the  edges 
of  the  boxes  tend  to  "  pinch  in  "  and  scrape  off  the  film  from  the 
journal.  To  prevent  this  these  edges  should  be  cut  away,  thus 
also  forming  an  excellent  oil  channel  for  longitudinal  distribution 
of  the  oil  where  the  pressure  is  least.  An  excellent  arrange- 
ment of  boxes  for  distributing  the  oil  and  maintaining  the  film 

*  Tower's  experiments,  Proc.  Inst.  M.  E.,  1883  and  1885.  See  further  Prof. 
Reynold's  paper  "  On  the  Theory  of  Lubrication,"  Phil.  Trans.,  1886,  Part  I, 
pp.  157-234. 

t  Professor  Reynolds  states,  in  Phil.  Trans.  1886,  Part  i,  p.  161,  that  if  viscosity 
were  constant  the  friction  would  be  inversely  proportional  to  the  difference  in  radii 
of  the  journal  and  the  bearing. 


204 


MACHINE  DESIGN. 


is  shown  in  Fig.  134  B,  which  is  copied  from  Vol.  27,  Trans. 
A.  S.  M.  E.,  p.  484.  With  true  film  lubrication  it  is  quite  defi- 
nitely established  that  the  coefficient  of  friction  varies  directly 
as  the  square  root  of  the  surface  velocity  and  inversely  as  the 

c  \/v 

unit  pressure,  so  that   n  = »    where   /*  =  coefficient  of  fric- 

P 

tion,  e  is  a  constant  depending  upon  the  lubricant  (being  .21 
for  rape  oil,  .29  for  lard  oil,  .32  for  mineral  oil  and  .43  for  mineral 


BOTTOM  HALF 


FIG.  1346. 

grease),  v  is  the  velocity  of  rubbing  in  feet  per  second,  and  p 
is  the  pressure  in  pounds  per  square  inch  of  projected  bearing 
area. 

With  pad  lubrication  or  where  the  oil  is  fed  drop  by  drop  there 
is  a  tendency  for  the  film  to  be  too  thin  or  to  break  down,  allow- 
ing contact  of  the  metallic  surfaces,  and  the  highly  favorable 
condition  of  fluid  friction  disappears.  The  conditions  then  lie 
between  "  fluid  friction  "  and  "  solid  friction  "  and  are  too  com- 
plex for  the  statement  of  consistent  results  but  it  may  be  approxi- 


JOURNALS,  BEARINGS,   AND  LUBRICATION.  205 

mately  stated  that,  with  good  pad  lubrication,  the  coefficient 
of  friction  will  be  about  twice  that  of  film  lubrication.  With 
drop  by  drop  lubrication  the  value  of  the  coefficient  may  become 
anything  between  twice  that  for  film  lubrication  (i.e.  =  .01), 
and  o.i 8,  the  value  determined  by  Morin  for  dry  journals.  It 
1  ecomes  apparent  that  some  system  of  forced  or  flooded  lubri- 
cation whereby  a  continuous  film  is  insured  is  of  utmost  value  in 
maintaining  efficiency. 

Let  /,  Fig.  135,  represent  a  journal  with  its  box,  and  let  A, 
B,  and  C  be  oil-holes.      If  oil  is  introduced  into  the  hole  A,  it 


FIG.  135. 

will  tend  to  flow  out  from  between  the  rubbing  surfaces  by  the 
shortest  way,  i.e.,  it  will  come  out  at  D.  A  small  amount 
will  probably  go  toward  the  other  end  of  the  box  because  of 
capillary  attraction,  but  usually  none  of  it  will  reach  the  middle 
of  the  box.  If  oil  be  introduced  at  C,  it  will  come  out  at  E.  A 
constant  feed,  therefore,  might  be  maintained  at  A  and  C,  and 
yet  the  middle  of  the  box  might  run  dry.  If  the  oil  be  introduced 
at  B,  however,  it  tends  to  flow  equally  in  both  directions,  and 
the  entire  journal  is  lubricated.  The  conclusion  follows  that 
oil  ought,  when  possible,  to  be  introduced  at  the  middle  of  the 
length  of  a  cylindrical  journal.  It  should  be  introduced  as  far 
as  possible  from  the  side  where  the  forces  press  the  journal  and 
box  closest  together.*  If  a  conical  journal  runs  at  a  high  velocity, 
the  oil  under  the  influence  of  centrifugal  force  tends  to  go  to 
the  large  end  of  the  cone,  and  therefore  the  oil  should  be  intro- 
duced at  the  small  end  to  insure  its  distribution  over  the  entire 
journal  surface. 

*  Tower'.;  cx:>e  ."merit;,  Proc.  Inst.  l.~.Y,.:  i33;  end  iG'>-. 


2C6 


MACHINE  DESIGN. 


If  the  end  of  a  vertical  thrust-journal  whose  outline  is  a 
cone  or  tractrix,  as  in  Fig.  136,  dips  into  a  bath  of  oil,  B,  the 
oil  will  be  carried  by  its  centrifugal  force,  if  the  velocity  be 
high,  up  between  the  rubbing  surfaces,  and  will  be  delivered 
into  the  groove  A  A.  If  holes  connect  A  and  B,  gravity  will 
return  the  oil  to  B,  and  a  constant  circulation  will  be  main- 
tained. If  the  thrust-journal  has  simply  a  flat  end,  as  in  Fig. 
137,  the  oil  should  be  supplied  at  the  center  of  the  bearing; 
centrifugal  force  will  then  distribute  it  over  the  entire  surface. 
If  the  oil  is  forced  in  under  a  pressure  sufficient  to  "float"  the 
shaft  the  friction  will  be  greatly  reduced.  Vertical  shaft  thrust- 
journals  may  usually  be  arranged  to  run  in  an  oil-bath.  Marine 
collar  thrust- journals  are  always  arranged  to  run  in  an  oil-bath. 


FIG.  136. 


FIG  137. 


FIG.  138. 


Sometimes  a  journal  is  stationary,  and  the  box  rotates 
about  it,  as  in  the  case  of  a  loose  pulley,  Fig.  138.  If  the  oil 
is  introduced  into  a  tube,  as  is  often  done,  its  centrifugal 
force  will  carry  it  away  from  the  rubbing  surface.  But  if  a 
hole  is  drilled  in  the  axis  of  the  journal,  the  lubricant  intro- 
duced into  it  will  be  carried  to  the  rubbing  surfaces  as  required. 
If  a  journal  is  carried  in  a  rotating  part  at  a  considerable  dis- 
tance from  the  axis  of  rotation,  and  it  requires  to  be  oiled  while 
in  motion,  a  channel  may  be  provided  from  the  axis  of  rota- 


JOURNALS,  BEARINGS,   AND    LUBRICATION. 


207 


tion,  where  oil  may  be  introduced  conveniently,  to  the  rub- 
bing surfaces,  and  the  oil  will  be  carried  out  by  centrifugal 
force.  Thus  Fig.  139  shows  an  engine-crank  in  section.  Oil 
is  introduced  at  b,  and  centrifugal  force  carries  it  through 
the  channel  provided  to  a,  where  it  serves  to  lubricate  the  rub- 


FIG.  139. 


FIG.  140. 


bing  surfaces  of  the  crank-pin  and  its  box.  If  a  journal  is 
carried  in  a  reciprocating  machine  part,  and  requires  to  be 
oiled  while  in  motion,  the  "  wick-and- wiper  "  method  is  one  of 
the  best.  (See  Fig.  140.)  An  ordinary  oil-cup  with  an  adjust- 
able feed  is  mounted  in  a  proper  position  opposite  the  end  of 
the  stroke  of  the  reciprocating  part,  and  a  piece  of  flat  wick 
projects  from  its  delivery-tube.  A  drop  of  oil  runs  down  and 
hangs  suspended  at  its  end.  Another  oil  cup  is  attached  to 
the  reciprocating  part,  which  carries  a  hooked  "wiper,"  C. 
The  delivery-tube  from  C  leads  to  the  rubbing  surfaces  to  be 
lubricated.  When  the  reciprocating  pait  reaches  the  end  of 
its  stroke  the  wiper  picks  off  the  drop  of  oil  from  the  wick 
and  it  runs  down  into  the  oil-cup  C,  and  thence  to  the  sur- 
faces to  be  lubricated.  This  method  applies  to  the  oiling  of 
the  cross-head  pin  of  a  steam  engine.  The  same  method  is 
sometimes  applied  to  the  crank-pin,  but  here,  through  a  part 


208 


MACHINE   DESIGN. 


of  the  revolution,  the  tendency  of  the  centrifugal  force  is  to 
force  the  oil  out  of  the  cup,  and  therefore  the  plan  of  oiling 
from  the  axis  is  probably  preferable. 

When  journals  are  lubricated  by  feed- oilers,  and  are  so 
located  as  not  to  attract  attention  if  the  lubrication  should  fail 
for  any  reason,  "  tallow-boxes  "  or  "  grease-cups  "  are  used. 
These  are  cup-like  depressions  usually  cast  in  the  box-cap 
and  communicating  by  means  of  an  oil-hole  with  the  rubbing 
surface.  These  cups  are  filled  with  grease  that  is  solid  at 
the  ordinary  temperature  of  the  box,  but  if  there  is  the  least 
rise  in  temperature  because  of  the  failure  of  the  oil-supply, 
the  grease  melts  and  runs  to  the  rubbing  surfaces,  and  sup- 
plies the  lubrication  temporarily.  This  safety  device  is  used 
very  commonly  on  line-shaft  journals. 

The  most  common  forms  of  feed-oilers  are:  I.  The  oil-cup 
with  an  adjustable  valve  that  controls  the  rate  of  flow.  II.  The 
oil-cup  with  a  wick  feed  (Fig.  141).  The  delivery  has  a  tube 


FIG.  141. 


FIG.  142. 


FIG.  143. 


inserted  in  it  which  projects  nearly  to  the  top  of  the  cup.  In 
this  tube  a  piece  of  wicking  is  inserted,  and  its  end  dips  into 
the  oil  in  the  cup.  The  wick,  by  capillary  attraction,  carries 
the  oil  slowly  and  continuously  over  through  the  tube  to  the 


JOURNALS,  BEARINGS,  AND  LUBRICATION.  209 

rubbing  surfaces.  III.  The  cup  with  a  copper  rod  (Fig.  142;. 
The  oil-cup  is  filled  with  grease  that  melts  with  a  very  slight 
elevation  of  temperature,  and  A  is  a  small  copper  rod  dropped 
into  the  delivery-tube  and  resting  on  the  surface  of  the  journal. 
The  slight  friction  between  the  rod  and  the  journal  warms 
the  rod  and  it  melts  the  grease  in  contact  with  it,  which  runs 
down  the  rod  to  the  rubbing  surface.  IV.  Sometimes  a  part 
of  the  surface  of  the  bottom  half  of  the  box  is  cut  away  and 
a  felt  pad  is  inserted,  its  bottom  being  in  contact  with  an  oil- 
bath.  This  pad  rubs  against  the  surface  of  the  journal,  is 
kept  constantly  soaked  with  oil,  and  maintains  lubrication. 

Ring-and-chain  lubrication  may  be  considered  as  special 
forms  of  bath  lubrication.  Fig.  143  shows  a  ring  oiling  bearing. 
A  loose  ring  rests  on  top  of  the  journal,  the  upper  box  being 
cut  away  to  permit  this;  the  ring  surrounds  the  lower  box 
and  extends  into  a  reservoir  filled  with  oil.  The  rotation  of 
the  shaft  carries  the  ring  with  it,  which,  in  turn,  brings  up  a 
constant  supply  of  oil  from  the  reservoir.  The  annular  spaces 
A- A  catch  all  oil  which  works  out  along  the  shaft  and  return 
it  to  the  reservoir. 

Graphite  is  winning  a  deservedly  high  place  as  a  lubricant  for 
certain  conditions.  Its  action  is  to  reduce  "solid  friction"  by 
filling  the  inequalities  in  the  surfaces  of  the  relatively  moving 
members,  giving  each  a  smooth,  slippery  coating,  thereby  reduc- 
ing the  coefficient  of  friction.  It  is  particularly  useful  when  the 
conditions  of  pressure  or  temperature  are  such  as  would  tend 
to  squeeze  out,  gum,  or  destroy  liquid  lubricants,  if  these  were 
used  alone. 

Although  it  may  be  applied  in  some  cases  in  dry  flake  form, 
it  is  customary  to  use  it  in  the  form  of  a  mixture  with  oils,  grease, 
or  even  water.  Caution  must  be  observed  that  the  graphite  used 
is  free  from  all  grit. 


CHAPTER  XIII. 

ROLLER-  AND   BALL-BEARINGS. 

141.  General  Considerations. — By  substituting  rolling  motion 
in  bearings  in  place  of  relative  sliding,  friction  losses  can  be 
greatly   reduced.     In   the   design   of  such   bearings   there   are 
four  points  to  be  borne  in  mind : 

I.  The   arrangement   of   the  parts  and  their   form  must  be 
such  that  their  relative  motion  is  true  rolling  with  the   least 
possible  amount  of  sliding. 

II.  The  form  of  the  constraining  surfaces  must  be  such  that 
the  rolling  parts  will  not  have  any  effective  tendency  to  leave 
their  proper  guides  or  "races." 

III.  The  rollers  and  balls  must  not  be  unduly  loaded. 

IV.  Provision  must  be  made  to  admit  the  lubricant,  and  to 
exclude  all  dust  and  grit. 

These  points  will  be  considered  in  the  order  given. 

142.  I.  Rolling,  Sliding,  and  Spinning.     (See  Fig.  144.) — At 


A  is  shown  the  longitudinal  section  of  a  cylindrical  ball-bearing 
of  the  simplest   form  stripped   of  all  auxiliary   parts.      At  B 


ROLLER-  AND  BALL-BEARINGS. 


211 


is  shown  the  same  for  a  roller-bearing.  At  C  is  a  cross-section 
of  either,  showing  but  one  ball  or  roller  R.  S  is  the  journal 
and  T  the  box.  Consider  T  as  stationary,  then  the  point  of 
contact  of  R  and  S  would  have  the  same  motion  relative  to  T 
whether  considered  as  a  point  of  R  or  of  S,  and  if  the  surface 
friction  were  sufficient  there  would  be  no  reason  for  slippage. 
As  a  matter  of  fact,  in  the  actual  bearing  there  will  be  a  slight 
amount  of  slipping  .at  both  of  the  points  of  contact.  This  form 
of  bearing  is  called  the  "  two-point  bearing,"  because  there  are 
two  points  of  contact.  All  cylindrical  roller-bearings  are  of 
this  fundamental  form.  In  order  to  have  them  of  practical  use 
the  rollers  must  be  held  in  a  case  or  "cage"  so  that  their  axes 
will  always  remain  parallel  with  the  axis  of  the  shaft.  Fig.  145 
shows  such  a  "cage"  with  rollers  in  place. 


FIG.  145. 

Since  the  rollers  are  generally  of  hardened  and  ground  steel 
the  best  service  with  the  least  wear  will  be  given  when  the 
engaging  surfaces  are  of  the  same  material.  To  meet  this  when 
the  shaft  is  of  soft  steel,  say,  and  the  box  of  cast  iron,  a  hardened 
and  ground-steel  ring  is  fitted  over  the  shaft  as  a  shell  and 
another  inside  the  box  as  a  bushing,  and  the  rollers  run  between 
the  outer  surface  of  the  former  and  the  inner  surface  of  the 
latter. 

Ball-bearings  are  subject  to  an  action  known  as  "spinning." 
To  illustrate  this,  consider  the  three-point  ball-bearing  shown 


MACHINE  DESIGN. 


in  Fig.  146.  Here  the  centres  are  as  shown  in  B,  and  the  con- 
ditions are  correct  for  theoretical  rolling  as  long  as  point  contact 
is  maintained  and  axis  C-D  remains  parallel  to  axis  E-F.  But 
when  the  bearing  is  in  use  the  points  of  contact,  on  each  side 
of  R,  with  T  become  small  areas,  as  shown  in  B.  Considering 
the  relative  motion  of  R  and  T  at  any  instant  it  will  be  seen  that 


FIG.  146. 


FIG.  147. 


there  is  an  action  on  each  side  of  the  ball  akin  to  that  of  a  small 
thrust-bearing.  The  rubbing  produced  in  this  manner  naturally 
causes  undesirable  friction.  This  is  the  action  known  as  "spin- 
ning." 

Obviously  it  is  even  more  marked  in  the  case  of  a  four- 
point  bearing,  as  shown  in  Fig.  147. 

Here,  also,  there  is  pure  rolling  motion  as  long  as  point  con- 
tact is  maintained,  and  the  axes  C-D  and  E-F  remain  parallel 
to  axis  G-H;  but  as  soon  as  the  load  is  applied  the  points  of 
contact  become  areas,  and  "spinning"  results  at  four  surfaces. 
Experiments  bear  out  the  conclusion  that  a  properly  designed 
two-point  bearing  will  have  less  friction  than  a  three-point,  and 
a  three-point  will  have  less  than  a  four-point. 

In  a  "race  "  whose  radius  of  curvature  is  just  equal  to  that 
of  the  ball  the  friction  becomes  excessive.  Such  races  should 
never  be  used.  (See  Fig.  148.) 

A  force  acting  at 'the  surface  of  a  ball  will  tend  to  rotate 
it  about  an  axis  parallel  to  the  tangent  plane  in  which  the  actu- 


ROLLER.-  AND  BALL-BEARINGS. 


213 


ating  force  lies;  furthermore,  this  axis  will  be  at  a  right  angle 
with  the  direction  of  the  force.  This  is  true  because  it  is  merely 
a  special  application  of  the  general  law  that  a  force  applied  to 
a  body  will  tend  to  move  it  in  the  direction  of  action  of  the  force. 


FIG.  148. 


FIG.  149. 


The  general  law  for  the  form  of  rolling  bearings  may  now  be 
stated  as  follows: 

For  true  rolling,  the  constraining  surfaces  of  the  journal 
and  box  (i.e.,  the  "races")  must  be  so  formed  that  the  axes  of 
rotation  of  the  rollers  or  balls  will  all  intersect  the  main  axis 
of  the  bearing  at  a  fixed  point  throughout  the  complete  revolu- 
tion of  the  journal.  This  may  be  made  clear  by  examples. 

Fig.  149  shows  a  ball  or  roller  R  held  between  two  similar 
plates  T  and  S.  The  upper  plate,  T,  presses  down  on  R  with 
a  force  P  which  is  transmitted  through  R  to  5. 

By  the  principles  of  so-called  "  rolling  friction,"  to  roll 
T  on  R  will  require  a  force  }P  (i.e.,  proportional  to  P)  to 
overcome  the  resistance.  The  motion  of  T  on  R  causes  R 
to  roll  on  5,  to  which  rolling  there  is  induced  a  resistance  also 
equal  to  /P,  but  in  the  opposite  direction  as  regards  R.  These 
two  forces  being  equal,  opposite,  and  applied  at  the  same  dis- 
tance from  the  center  of  R,  form  a  couple  whose  effect  would 
be  to  give  R  a  motion  of  rotation  about  an  axis  through  its 
center,  and  perpendicular  to  the  plane  in  which  they  both  lie. 
This  case  is  similar  to  those  shown  in  Fig.  144,  except  that 


214 


MACHINE  DESIGN. 


in  the  cases  there  shown  5  and  T  are  not  plane  surfaces.  Each 
ball  in  case  A  and  each  roller  in  case  B  tends  to  rotate  about  an 
axis  (relatively  to  the  "  cage,'*  not  shown)  as  indicated  by  the 
dotted  lines.  In  both  cases  the  individual  axes  all  intersect 
the  main  axis  of  the  journal  at  a  fixed  point,  namely,  at  infinity, 
throughout  the  revolution.  The  general  law  for  true  rolling 
is  therefore  fulfilled. 

In  the  cases  shown  in  Figs.  146  and  147,  obviously  the 
same  conditions  hold. 

Next  consider  the  thrust- bearings  shown  in  Fig.  150: 


FIG.  150. 

Take  case  A  first.  5  is  the  moving  member,  T  the  sta- 
tionary member,  R  one  of  the  balls,  and  OF  is  the  axis  of 
rotation  of  S  relatively  to  T.  The  center  of  the  ball  is  at  any 
distance  r  from  the  axis  OF,  and  its  points  of  contact  with  S 
and  T  are  termed  A  and  B  respectively.  Relatively  to  the 
inclosing  cage  (not  shown)  all  parts  of  the  ball  in  obedience 
to  the  acting  force  tend  to  rotate  about  the  axis  OX  which 
always  cuts  OF  at  O.  It  is  not  essential  that  the  angle  XOY 
be  a  right  angle;  it  is  essential  that  a  line  joining  A  and  B 
should  pass  through  the  center  of  R  and  be  perpendicular  to  OX. 
It  is  obvious  that  the  conditions  for  true  rolling  are  fulfilled. 

As  OX  completes  one  revolution  about  OF  the  ball  will 
rotate  about  its  axis  A-B  relative  to  both  S  and  T.  Further- 
more, since  A  and  B  become  small  areas  under  the  action 
of  the  load,  it  is  impossible  to  avoid  the  undesirable  effect  of 
"spinning." 


ROLLER-  AND   BALL  BEARINGS.  215 

In  case  B,  as  S  rotates  relative  to  T,  the  point  D  common 
to  R  and  5  will  have  a  linear  velocity  proportional  to  r2  and, 
similarly,  C's  linear  velocity  will  be  proportional  to  r\.  If 
AD  and  BC  were  two  equal,  independent  circular  disks,  each 
would  have  true  rolling  motion,  and  BC  would  make  r\  revolu- 
tions, while  AD  would  make  r2.  But  BC  and  AD  are  both 
disks  of  the  same  roller,  R,  and  cannot  rotate  relative  to  each 
other;  hence  they  must  each  make  the  same  number  of  revo- 
lutions, and  points  C  and  D  of  the  disks  would  have  to  have 
the  same  velocity,  which  is  inconsistent  with  the  conditions  of 
motions  of  C  and  D  as  points  of  S.  Hence  a  roller  cannot 
be  correctly  used  for  a  thrust-bearing.  Short  rollers  securely 
held  in  cages  are  used  in  practice,  but  experiments  show  that 
they  are  not  as  efficient  as  properly  designed  forms.* 

Consider  case  C.  Relative  to  T,  the  double  point  D  will 
have  a  linear  velocity  proportional  to  r2  and  C  will  have  a 
linear  velocity  proportional  to  r\.  Consider  AD  and  BC  as 

T>f*         „ 

independent  disks  so  proportioned  that  -rj:  =  —  .     If  D  has  a 

A.L)     T2 

linear  velocity  proportional   to    r2,  then   the    angular   velocity 

?2 

oi  AD   about    its    axis   OX  will   be    proportional   to    —  r. 


Similarly,  the  angular  velocity  of    BC  about  axis  OX  will  be 
proportional  to 


Angular  velocity  of  AD      nAD      BC  r2     f\  TZ 
Angular  velocity  of  BC        r\        AD  r\     r2  r\ 
~ 


BC    ri 

since  -r-jFi  =~~- 

AD    r-2, 

Hence  the  disks  AD  and  BC  have  the  same  angular  velocity 
about  the  axis  OX,  and  may  form  parts  of  the  same  body. 

*  See  article  by  T.  Hill  in  American  Machinist,  Jan.  5,  1899.     Also  description 
of  bearing  by  C.  R.  Pratt,  same  periodical,  June  27,  1901. 


2l6  MACHINE  DESIGN. 

This  will  be  true  of  any  pair  of  disks  of  the  cone  OBC.  Any 
frustum  of  a  cone  whose  apex  lies  anywhere  on  the  axis  OF 
will  therefore  fulfill  the  conditions  for  true  rolling  motion  rela- 
tively to  T  when  actuated  by  S. 

In  each  of  the  foregoing  cases  the  rolling  members  must  be 
held  in  suitable  "  cages,"  or  they  will  yield  to  the  tendency  to 
displace  them. 

In  ball  thrust-bearings  it  is  desirable  to  so  arrange  the  balls 
in  the  cage  that  each  one  will  have  a  separate  path,  as  this 
minimizes  wear. 

For  a  three-point  thrust  ball-bearing  the  form  of  the  races 
to  permit  true  rolling  must  be  as  shown  in  Fig.  151  to  be  in 
accordance  with  the  principles  just  demonstrated.  The  groove- 
angle  should  be  as  flat  as  possible  to  reduce  the  friction  effect 
of  "  spinning." 

About  120°  will  be  found  a  good  practicable  value. 

The  ball  becomes  akin  to  a  cone  as  far  as  its  relation  with  7\ 
with  which  it  has  two  points  of  contact,  is  concerned.  It  remains 
as  a  ball  as  far  as  its  relation  with  S  is  concerned.  In  each, 
case  the  motion  imparted  to  the  ball  tends  to  rotate  it  about  the 
correct  axis  OX  and  the  conditions  for  true  rolling  are  sat- 
isfied. The  sides  of  the  race  are  tangent  to  the  ball  where  it 
is  cut  by  any  line  A-B  which  passes  through  O. 

A  four-point  ball-bearing  must  be  designed  according  to 
the  principles  indicated  in  Fig.  152  for  true  rolling  motion. 
As  far  as  its  motion  relations  with  T  and  5  are  concerned,  the 
ball  becomes  akii  to  a  cone.* 

*  The  method  of  laying  out  the  groove  in  Fig.  151  is  as  follows:  —  The  axes  of 
rotation  of  the  balls  cut  the  main  axis  of  the  bearing  at  O.  Draw  the  lower  surface 
of  5  tangent  to  the  ball  at  C  and  parallel  to  the  ball  axis  OX.  Draw  the  line  OB 
cutting  the  ball  at  A  and  B,  and  draw  tangent  surfaces  normal  to  the  radii  of  the 
ball  at  A  and  B.  These  surfaces  form  the  groove  angle  BDA.  If  the  first  trial 
gives  too  sharp  a  groove  angle,  increase  the  angle  XOB  and  repeat  the  construction. 
If  BDA  is  too  flat,  decrease  XOB. 

For  the  four-point  bearing  shown  in  Fig.  15.2  the  same  method  is  used  for  deter- 
mining the  groove  in  5  as  well  as  T. 


ROLLER-  AHD  BALL-BEARINGS. 


217 


Similarly  a  cup  and  cone  three-point  bearing  should  have 
the  form  shown  in  Fig.  153. 


FIG.  153.  FIG.  152. 

143.  II.  The  form  of  the  constraining  surfaces  must  be  such 
in  ball-bearings  that  the  balls  will  not  have  any  effective  tendency 
to  leave  their  proper  paths.  The  use  of  cages  for  this  purpose 
has  already  been  mentioned. 

If  two-point  bearings  without  cages  are  desired,  the  section 
of  each  race  should  be  the  arc  of  a  circle  whose  radius  is  TV 
to  f  of  the  diameter  of  the  ball.  In  two-point  bearings  the 
points  of  pressure  must  always  be  diametrically  opposite. 

In  three-  and  four-point  bearings  where  the  races  are  properly 
formed  for  true  rolling,  as  explained  in  the  preceding  section,  the 
tendency  for  the  balls  to  leave  the  races  is  reduced  to  a  minimum. 

One  point,  however,  needs  further  consideration.  In  cup- 
and  cone-bearings  it  is  impossible  to 
keep  a  tight  adjustment  at  all  times, 
and  the  least  play  will  allow  some  of 
the  balls  on  the  unloaded  side  of  the 
bearing  to  get  out  of  place. 

Fig.  154  shows  such  a  bearing 
loosely  adjusted. 

The  loaded  cup  is  forced  down  so 
that  its  axis  lies  below  the  axis  of 
the  cone.  The  top  ball  is  held  correctly  in  place  for  true 


2i8  MACHINE  DESIGN. 

rolling;  the  lower  ball  is  free  to  roll  to  one  side  as  seen. 
Investigation  has  shown  that  the  angles  a  and  /?  should  each 
be  at  least  as  great  as  25°  in  order  to  return  the  displaced  ball 
easily  to  its  proper  path  by  the  time  it  becomes  subjected  to 
the  load.  If  the  angles  are  too  acute  there  is  a  tendency  for 
the  balls  to  wedge  in  their  incorrect  positions,  causing  rapid 
wear  or  even  crushing.* 

144.  III.  Allowable  Loading.  —  Careful  experiments  show 
that  for  high  efficiency  and  durability  the  loads  on  balls  and 
rollers  should  be  very  much  less  than  they  could  be  with  safety 
as  far  as  their  strength  is  concerned.  f 

Let  PQ  equal  the  load  in  pounds  which  is  allowable  for  a  single 
ball  or  roller.  Then  for  balls 

Po  =  Kd?, 
d=  diameter  of  ball  in  inches. 

K  =  1  500  for  hardened  steel  balls  and  races,  two-point  bearing, 

with  circular-arc  races  having  radii  equal  to  f  d. 
K=   750  for  hardened   steel   balls  and    races,  three-  and  four- 

point  bearing. 

For  two-point  bearing  with  flat  races,  K  =  500. 
For  cast-iron  balls  and  races  use  two-fifths  of  these  values. 
For  rollers  P0=Kdl. 

d  =  diameter  of  roller  in  inches  =  mean  diameter  of  cone, 
/  =  length  of  roller  in  inches, 
K  =  4oo  for  cast  iron, 
K  =  IOCQ  for  hardened  steel. 
In  thrust-bearings,  if  the  total  load  =  P  and  the  number  of 

p 

balls  =  w,  we  have  for  either  balls  or  rollers  PQ  =  —. 

n 


*  See  article  by  R.  Janney  in  American  Machinist,  Jan.  5,  1899. 
t  See  excellent  article  by  Professor  Stribeck  in  Z.  d.  V.  d.  L,  Jan.  19  and  26, 
1901. 


ROLLER- AND  BALL-BEARINGS.  2X9 

In  cylindrical  bearings  the  load  is  always  greatest  at  one 
side  of  the  bearing,  the  balls  or  rollers  on  the  opposite  side 
being  entirely  unloaded.  It  has  been  found  that  the  load  on 

the  heaviest  loaded   ball  or  roller =P0  =  —P,  where  n  is  the 

n 

number  of  balls.* 

145.  Size   of  Bearing. — To  determine  the  size  of  the  ball 
circle   (i.e.   the   middle  diameter  of   the 
61  race  " )  given   the   number  of   balls   n 
and  their  diameter  d.     (See  Fig.  155.) 

r  =  radius  of  ball.  R  =  radius  of  ball 
circle.  Join  the  centers  of  two  consec- 
utive balls  by  the  chord  AB  =  2r.  From 
the  center  of  the  ball  circle,  O,  draw  two 
radii,  one  to  A  and  the  other  to  the  FlG'  I55' 

mid-point  of  A-B.     Call  the  angle  included  between  the  radii  a. 
Then 

r  =  R  sin  a,  and, 

180° 

since  a= , 

n 


.    i8o°' 
sin  — 
n 


*  Mr.  Henry  Hess,  of  the  Hess-Bright  Mfg.  Co.,  in  a  letter  to  the  authors  says:  — 
"It  is  a  fact,  that  has  been  determined  by  experience,  that  in  radia  [i.e.,  cylindrical] 
ball  bearings  the  speed  has  very  little  influence  within  very  wide  limits.  In  my 
practice  ...  I  pay  no  attention  to  speed  of  radial  bearings  up  to  3000  rpm  as 
influencing  the  load  so  long  as  such  speed  is  fairly  uniform  and  so  long  as  the  load 
is  fairly  uniform.  When  neither  speed  nor  load  are  uniform  the  percussive  effect 
of  rapid  changes  must  be  taken  in  consideration;  unfortunately,  so  far  at  least,  the 
factors  are  entirely  empirical  and  allowances  are  made  by  a  comparison  with 
analogous  cases  of  previous  practice. 

The  case  is  different  with  thrust  bearings.  In  these,  speed  is  a  very  decided 
factor  in  the  carrying  capacity  even  though  speed  and  load  be  uniform.  Here 
again  no  rational  formula  has  yet  been  developed  to  adequately  represent  the 
different  elements,  but  carrying  capacities  for  different  speeds  of  standard  bearings 


MACHINE  DESIGN. 


This  is  the  radius  of  a  circle  on  which  the  centers  of  the 
balls  will  lie  when  their  surfaces  are  all  in  contact.  It  is  desir- 
able to  allow  some  clearance  between  the  balls.  This  may 
be  as  much  as  0.005  incn  between  each  pair  of  balls  provided 


the  total  allowance  does  not  exceed  — . 

4 


When  the  total  clear- 


ance has  been  decided  upon,  it  may  be  allowed  for  by  making 
the  actual  radius  of  the  ball  circle  larger  than  R  by  an  amount 
one  sixth  of  the  total  clearance  desired. 

146.  IV.  Lubrication  and  Sealing. — On  account  of  "spin- 
ning," faulty  adjustment,  and  unavoidable  slippage,  rolling 
bearings  should  be  properly  lubricated.  As  they  are  extremely 
sensitive  to  the  presence  of  dust  and  grit,  care  must  be  exer- 
cised that  the  lubricant  be  admitted  without  any  danger  of 
the  entrance  of  these. 

Sealed  oil-holes,  dust-caps,  and  felt  washers  are  commonly 
used  both  to  retain  the  lubricant  for  bath  lubrication  and  for 
keeping  out  all  dirt. 

have  been  experimentally  determined,  since  it  was  quite  feasible  to  get  different 
uniform  speeds  and  determine  under  what  load  the  carrying  capacity  was  reached. 
We  found,  for  instance,  that  for  a  thrust  bearing  employing  18  —  \"  balls,  the 
permissible  load  at  10  rpm  was  2400  pounds;  at  300  rpm  —  650  pounds;  at  1000  — 
450  pounds;  and  at  1500  only  330  pounds.  We  also  found  that,  generally  speaking, 
it  was  inadvisable  to  use  this  type  of  bearing  for  speeds  materially  above  1500  rpm." 
An  analysis  of  certain  standard  thrust  bearings  in  connection  with  the  makers' 
catalog  allowances  for  loads,  gives  at  various  speeds :  — 

LOAD  PER  BALL,  POUNDS. 


R.  P.  M. 

1/4"  Ball. 

5/1  6"  Ball. 

3/8"  Ball. 

1500 

27.8 

35-4-44-7 

80.7 

1000 

33-3 

4i-7-57-i 

91.7 

500 
300 
150 

10 

41.7 
55-6 
61.2 

210 

52.1-71.4 
66.7-89.2 
83.3-107 
229-339 

129 
153 
193 
560 

C.  R.  Pratt,  Trans.  A.S.M.E.,  Vol.  27  gives  as    limiting  load  per  \"  ball,  100 
pounds  at  700  rpm  with  a  6"  diameter  circle  of  rotation. 


CHAPTER  XIV. 

COUPLINGS  AND  CLUTCHES. 

147.  Couplings   and  Clutches  Defined. — Couplings  are  those 
machine  parts  which  are  used  to  connect  the  ends  of  two  shafts 
or  spindles  in  such  a  manner  that  rotation  of  the  one  will  pro- 
duce an  identical  rotation  of  the  other.     They  are  therefore 
in  the  nature  of  fastenings,  and  may  be  classified  as  perma- 
nent or  disengaging.    The  latter  are  frequently  called  clutches. 

148.  Permanent     Couplings. — The    simplest    form    of    per- 
manent  coupling  is   shown  in  Fig.  156,  and   is    known  as  the 
"  sleeve  "  or  "  muff  "  coupling.     Each  shaft  has  a  keyway  cut 


FIG.  156. 

at  the  end.  The  cast-iron  sleeve  of  the  proportions  indicated  is 
bored  an  exact  fit  for  the  shafts  and  has  a  keyway  cut  its  entire 
length.  When  the  sleeve  is  slipped  over  the  ends  of  the  shafts, 
the  key  is  driven  home  and  all  relative  rotation  is  prevented. 
The  key  may  be  proportioned  according  to  the  rules  laid  down 
in  the  chapter  on  Means  for  Preventing  Relative  Rotation. 


MACHINE  DESIGN. 


149.  Flange  couplings  are    frequently    used,   and   Fig.    157 
illustrates  the  type.    Approximate  proportions  are    indicated. 


FIG.  157. 

The  number  of  bolts  n  may  be  from  3  +  0.5^  to  3  +  ^.  Their 
diameter  d'  must  be  such  that  their  combined  strength  to  resist 
a  torsional  moment  about  the  axis  of  the  shaft  will  be  equal 
to  the  torsional  strength  of  the  shaft, 


/•=  allowable  'stress   in   outer   fiber  of  shaft,   pounds   per 

square  inch; 

d=  diameter  of  shaft,  inches; 
R  =  radius  of  bolt  circle,  inches  ; 
n  =  number  of  bolts  ; 
d'=  diameter  of  bolts,  inches; 

]»  =  allowable  shearing  stress  in  bolts,  pounds  per  square  inch. 
This  equation  will  approximately  give 


150.  Compression    couplings  of  three  forms  are  shown  in 
Figs.  158,  159,  and   160.     The  first  is  similar  to  the  ordinary 


COUPLINGS  AND  CLUTCHES. 


223 


sleeve  coupling  except  that  the  sleeve  is  split  in  halves  longi- 
tudinally and  is  tapered  toward  each  end  on  the  outside.  The 
two  rings  shown  are  driven  or  shrunk  on  these  tapers. 


FIG.  158. 

Instead  of  being  held  by  rings  the  half  sleeves  are  some- 
times bolted  together  as  shown  in  Fig.  159. 


FIG.  159. 

151.  The  "  Sellers "  coupling  is  shown  in  Fig.  160.  An 
outer  sleeve  A  is  bored  tapering  from  each  end.  A  split  cone 
bushing  B  is  inserted  at  each  end.  Openings  are  left  for  three 


bolts  by  means  of  which  B  and  B  are  drawn  toward  each  other 
and  thus  closed  down  on  the  shaft  with  great  force.  A  key 
is  also  provided. 


224 


MACHINE  DESIGN. 


CD? 


152.  Oldham's  Coupling. — For  all  of  the    foregoing  coup- 
lings the  axes  of  the  two  shafts  must  be  identical.     Where  the 

— h  axes   are   parallel,    Oldham's    coupling 

may  be  used.  It  is  shown  in  Fig.  161, 
and  consists  of  three  parts.  Each  shaft 
end  has  keyed  to  it  a  flange  or  disk 
which  has  a  diametral  groove  cut  across 
its  face.  Between  these  two  disks  is 
a  third  which  has  a  tongue  on  each  face.  The  tongues,  which 
just  fit  the  grooves  freely,  run  diametrically  across  the  disk 
and  are  at  a  right  angle  with  each  other. 

153.  Hooke's    Coupling. — For  axes  which  intersect,  Hooke's 
coupling  or  "  universal  joint  "  may  be  used.     It  is  shown  in 
outline  in  Fig.  162.     Each  shaft, has  a  stirrup  keyed  to  its  end. 


FIG.  161. 


FIG.  162. 

Each  stirrup  is  connected  by  turning  pairs  to  a  cross-shaped 
intermediate  member,  the  axes  of  whose  turning  pairs  are 
at  a  right  angle.  For  a  mathematical  analysis  of  this  mechanism 
the  reader  is  referred  to  Professor  Kennedy's  "Mechanics  of 
Machinery." 

154.  Flexible  Couplings. — Where  shafts  which  are  or  which 
may  become  slightly  out  of  alignment  are  to  be  connected, 
some  form  of  flexible  coupling  is  advisable.  Their  principle 
is  illustrated  in  Fig.  163.  Each  shaft  has  keyed  to  its  end  a 


COUPLINGS  AND   CLUTCHES. 


225 


disk  which  has  set  in  its  face  a  number  of  pins.  The  pins  are 
so  placed  that  those  in  the  one  circle  will  not  strike  those  in 
the  other  if  either  shaft  is  rotated  while  the  other  remains  at 
rest.  When  one  shaft  is  to  drive  the  other,  short  belts  are 
placed  on  the  pins  as  shown  in  B,  Fig.  163. 


FIG.  163. 

155.  Disengaging  couplings  are  of  two  general  classes: 
positive  drive  and  friction  drive.  Positive-drive  couplings  are 
commonly  called  toothed  or  claw  couplings.  They  consist 
of  two  members  having  projections  on  their  faces,  as  shown 
in  Fig.  164,  which  interlock  when  in  action.  A  is  keyed  rigidly 


r— 

' 

— 

'  — 

1 

I  

--1- 

jsE: 

- 

— 

-F— 

1  — 

—  i 

A     1  

JB 

— 

FIG.  164. 

to  its  shaft.     B  can  slide  along  its  shaft  guided  by  the  feather,  F, 
but  cannot  rotate  except  with  the  shaft.     When  B  is  moved 


226  MACHINE  DESIGN. 

in  the  direction  of  the  arrow,  its  teeth  engage  with  those  of  Ay 
and  the  two  shafts  must  have  the  same  motion  of  rotation. 
D  shows  the  end  of  the  lever  which  moves  B.  The  split  ring 
is  bolted  around  B  in  the  groove  E,  which  it  fits  freely.  Some- 
times short  blocks  which  fit  E  are  used  in  place  of  the  entire 
ring.  C  shows  an  end  view  of  the  half  clutch. 

Various  forms  of  teeth  may  be  used.*  If  n  equals  the  number 
of  teeth  and  R  equals  the  mean  radius  of  the  clutch  tooth  the 
following  equations  may  be  written  : 


nd3 
and  .-- 


In   the   first   equation,   fs~7~,   the   torsional  strength  of  the 

shaft,  is  equated  to  the  crushing  resistance  of  all  the  teeth 
opposed  to  the  torsional  stress.  A'  is  the  area  of  the  en- 
gaging face  of  one  tooth  and  }e  the  allowable  unit  crushing 
stress. 

In  the  second  equation  A  is  the  area  of  the  root  of  the 
tooth  subjected  to  shear  and  //  is  the  allowable  unit  shearing 
stress. 

156.  Friction  couplings  generally  consist  of  two  parts,  a 
hollow  cone,  A,  keyed  rigidly  to  one  shaft,  and  a  sliding  cone, 
B,  held  by  a  feather  on  the  second  shaft  as  seen  in  Fig.  165. 
By  means  of  a  lever,  B  can  be  forced  against  A  with  a  con- 
siderable axial  pressure.  This  induces  friction  between  the 
conical  surfaces,  which  friction  resists  relative  rotation.  To 
analyze  the  forces,  consider  Fig.  166  which  shows  two  conical 

*  See  article  in  American  Machinist,  July  9,  1903. 


COUPLINGS  AND   CLUTCHES. 


227 


surfaces  pressed  together  by  an  axial  force  P.    The  angle  at  the 
vertex  of  the  cone  is  20.. 


FIG.  165. 


FIG.  1 66. 


Let  the  coefficient  of  friction  =  /*  =  tan  <£. 
The  mean  cone- radius  =r. 

Total  pressure  between  the  two  surfaces  for  impending 
slippage  =2R. 

Total  normal  pressure  between  two   surfaces  at  rest=2AT. 

Total  friction  =  F  =  fi2N. 

It  is  clear  from  the  figure  that  P  =  2R  sin  (a 

Also,  N  =R  cos  <f>. 

2N  sin  ( 

cos  </> 
P 
But  2N  = — ,  where  F  =  friction  force  between  the  surfaces. 

p 

.'.  P  = —  (sin  a  +  ft  cos  a). 

From  this  equation  and  the  fact  that  the  turning  moment, 
M  =Fr,  the  clutch  can  be  designed  to  transmit  the  desired  power. 

The  angle  a  should  lie  between  7^°  and  12%°  in  order  to  avoid 
"sticking  "  on  the  one  hand  and  too  sudden  seizure  on  the  other. 


228  MACHINE  DESIGN. 

TABLE  XVIII. 

p  =  o.io  to  0.15  for  cast  iron  on  cast  iron 
=  0.15  to  0.20  for  cast  iron  on  paper 
=  0.20  to  0.30  for  cast  iron  on  leather 
=  0.20  to  0.50  for  cast  iron  on  wood. 

157.  Weston  Friction  Coupling. — For  heavy   duty  the  prin- 
ciple of  the  Weston  friction  coupling,  as  shown  in  Fig.  167,  may 


FIG.  167. 

be  used.  The  sleeve  A  carries  two  feathers  on  which  a  number 
of  iron  rings  C  can  slide  but  not  rotate.  Similarly  the  hollow 
sleeve  B  is  provided  with  feathers  which  prevent  the  rotation 
of  the  wooden  rings  D,  while  not  interfering  with  their  sliding. 


FIG.  i 68. 


Let  there  be  n  iron  rings.  Then,  when  B  is  pressed  toward 
A,  there  will  be  friction  induced  on  2w  +  i  surfaces.  If  P  is 
the  axial  pressure  and  /i  the  coefficient  of  friction,  the  total 


COUPLINGS  AND  CLUTCHES.  229 

friction  .F  =  /tP(2w  +  i),  and  if  r  =  ihe  mean  radius  of  the  rings, 
the  moment  which  can  be  transmitted  =  M  =Fr. 

158.  A  combination  friction-  and  claw-clutch  is  shown  in 
Fig.  1 68.  To  start  the  driven  shaft,  B  is  forced  to  the  right, 
thus  bringing  the  friction  cones  into  action.  When  the  driven 
shaft  has  attained  its  proper  speed,  B  is  suddenly  shifted  to  the 
1-jft,  thus  causing  the  claw-clutch  to  engage,  which  gives  the 
advantage  of  positive  driving. 

Professor  Bach  has  shown  that  the  energy  lost  in  friction  in 
making  a  "  running  start"  with  a  friction  clutch  just  equals 
the  kinetic  energy  of  the  shaft  and  attached  parts  at  the  speed 
to  which  they  have  been  brought.  This  shows  clearly  the 
enormous  wear  and  tear  to  which  friction  clutches  are  ordinarily 
subjected,  even  when  designed  to  run  practically  without  slip  in 
the  course  of  operation,  and  explains  their  rapid  deterioration. 

For  power  house  purposes  very  satisfactory  magnetic  clutches 
have  been  devised. 


CHAPTER   XV. 

BELTS. 

159.  Transmission  of  Motion  by  Belts. — In  Fig.  169,  let  A 
and  B  be  two  cylindrical  surfaces,  free  to  rotate  about  their  axes; 
let  CD  be  their  common  tangent,  and  let  it  represent  an  inex- 
tensible  connection  between  the  two  cylinders.  Since  it  is 
inextensible,  the  points  D  and  C,  and  hence  the  surfaces  of  the 


FIG.  169. 

cylinders,  must  have  the  same  linear  velocity  when  A  is  rotated 
as  indicated  by  the  arrow.  Two  points  having  the  same  linear 
velocity,  and  different  radii,  have  angular  velocities  which  are 
inversely  proportional  to  their  radii.  Hence,  since  the  surfaces 
of  the  cylinders  have  the  same  linear  velocity,  their  angular 
velocities  are  inversely  proportional  to  their  radii.  This  is  true 
of  all  cylinders  connected  by  inextensible  connectors.  Suppose 
the  cylinders  to  become  pulleys,  and  the  tangent  line  to  become 
a  belt.  Let  C'D'  be  drawn;  this  becomes  a  part  of  the  belt 
together  with  the  portions  DED'  and  CFCf,  making  it  endless, 
and  rotation  may  be  continuous.  The  belt  will  remain  always 

tangent  to  the  pulleys,  and  will  transmit  such  rotation  that  the 

230 


BELTS.  231 

angular  velocity  ratio  will  constantly  be  the  inverse  ratio  of  the 
radii  of  the  pulleys. 

The  case  considered  corresponds  to  a  crossed  belt,  but  the 
same  reasoning  applies  to  an  open  belt.    (See  Fig.  170.)    A  and 


B  are  two  pulleys,  and  CDD'C'C  is  an  open  belt.  Since  the 
points  C  and  D  are  connected  by  a  belt  that  is  practically  inex- 
tensible,  the  linear  velocity  of  C  and  D  is  the  same;  therefore 
the  angular  velocities  of  the  pulleys  are  to  each  other  inversely 
as  their  radii.  If  the  pulleys  in  either  case  were  pitch  cylinders 
of  gears  the  condition  of  velocity  would  be  the  same.  In  the 
first  case,  however,  the  direction  of  motion  is  reversed,  while 
in  the  second  case  it  is  not.  Hence  the  first  corresponds  to 
gears  meshing  directly  with  each  other,  while  the  second  corre- 
sponds to  the  case  of  gears  connected  by  an  idler,  or  to  the  case 
of  an  annular  gear  and  pinion.  While  in  many  places  positive 
driving-gears  are  indispensable,  it  is  frequently  the  case  that  the 
relative  position  of  the  axes  to  be  connected  is  such  as  would 
demand  gears  of  inconvenient  or  impossible  proportions,  and 
belts  are  used  with  the  sacrifice  of  positive  driving. 

Of  course  it  is  necessary  that  a  belt  should  have  some  thick- 
ness; and,  since  the  center  of  pull  is  the  center  of  the  belt,  it 
is  necessary  to  add  to  the  radius  of  the  pulley  half  the  thickness 
of  the  belt.  The  motion  communicated  by  means  of  belting, 
however,  does  not  need  to  be  absolutely  correct,  and  therefore 
in  practice  it  is  usually  customary  to  neglect  the  thickness  of  the 


232  MACHINE  DESIGN. 

belt.     The  proportioning  of  pulleys  for  the  transmission  of  any 
required  velocity  ratio  is  now  a  very  simple  matter. 

160.  Illustration. — A  line-shaft  runs  150  revolutions  per  min- 
ute, and  is  supported  by  hangers  with  16  inches  "drop."  It  is 
required  to  transmit  motion  from  this  shaft  to  a  dynamo  to  run 
1800  revolutions  per  minute.  A  3o-inch  pulley  is  the  largest 
that  can  be  conveniently  used  with  i6-inch  hangers.  Let 
v  =  the  diameter  of  required  pulley  for  the  dynamo;  then  from 
what  has  preceded  #-1-30  =  150-^1800,  and  therefore  #  =  2.5 
inches.  But  a  pulley  less  than  4  inches  diameter  should  not  be 
used  on  a  dynamo.*  Suppose  in  this  case  that  it  is  6  inches. 
It  is  then  impossible  to  obtain  the  required  velocity  ratio  with  one 
change  of  speed,  i.e.,  with  one  belt.  Two  changes  of  speed 


FIG.  171. 

may  be  obtained  by  the  introduction  of  a  counter-shaft.  By 
this  means  the  velocity  ratio  is  divided  into  two  factors.  If 
it  is  wished  to  have  the  same  change  of  speed  from  the  line  shaft 
to  the  counter  as  from  the  counter  to  the  dynamo,  then  each 


velocity  ratio  would  be  V (1800 -1-150)  =V  12  =3.46.     But  this 
gives  an  inconvenient  fraction,  and  the  factors  do  not  need  to  be 

*  This  limiting  size  is  determined  mainly  by  considerations  of  thickness  of 
belt  required  to  transmit  the  energy,  its  durability,  and  its  efficiency.  (See  §  171.) 


BELTS.  233 

equal.  Let  the  factors  be  3  and  4.  (See  Fig.  171.)  A  repre- 
sents the  line-shaft,  B  the  counter,  and  C  the  dynamo-shaft. 
The  pulley  on  the  line-shaft  is  30  inches,  and  the  speed  is  to  be 
three  times  as  great  at  the  counter,  therefore  the  pulley  on  the 
counter  connected  with  the  line-shaft  pulley  must  have  a  diam- 
eter one  third  as  great  as  that  on  the  line-shaft  =  10  inches. 
The  pulley  on  the  dynamo  is  6  inches  in  diameter  and  the  counter- 
shaft is  to  run  one  fourth  as  fast  as  the  dynamo,  and  therefore 
the  pulley  on  the  counter  opposite  the  dynamo-pulley  must  be 
four  times  as  large  as  the  dynamo-pulley  =  24  inches. 

161.  A    belt    may  be  shifted   from  one  part  of  a  pulley  to 
anolher  by  means  of  pressure  against  the  side  which  advances 
towards    the    pulley.      Thus  if,  in  Fig.  172, 

the  rotation  be  as  indicated  by  the  arrow, 
and  side  pressure  be  applied  at  A,  the  belt 
will  be  pushed  to  one  side,  as  is  shown,  and 
will  consequently  be  carried  into  some  new 
position  on  the  pulley  further  to  the  left  as  it 
advances.  Hence,  in  order  that  a  belt  may 

maintain  its  position  on  a  pulley,  THE  CEN- 

FIG.  172. 
TER    LINE     OF    THE    ADVANCING    SIDE    OF  THE 

BELT     MUST     BE     PERPENDICULAR     TO     THE     AXIS    OF    ROTATION. 

When  this  condition  Is  fulfilled  the  belt  will  run  and  trans- 
mit the  required  motion,  regardless  of  the  relative  position  of 
the  shafts. 

162.  Twist   Belts. — In  Fig.  173  the  axes  AB    and  CD   are 
parallel  to  each  other,  the  above  stated  condition  is  fulfilled, 
and  the  belt  will  run  correctly;   but  if  the  axis  CD  were  turned 
into  some  new  position,  as  C'D',  the  side  of  the  belt  that  advances 
toward  the  pulley  E  from  F  cannot   have   its   center  line  in  a 
plane  perpendicular  to  the  axis,  AB,  and  therefore  it  will  run  off. 
But  if  a  plane  be  passed  through  the  line  CD,  perpendicular  to 
the  plane  of  the  paper,  then  the  axis  may  be  swung  in  this  plane 


234 


MACHINE  DESIGN. 


in  such  a  way  that  the  necessary  condition  shall  be  fulfilled,  and 
the  belt  will  run  properly.  This  gives  what  is  known  as  a 
"twist"  belt,  and  when  the  angle  between  the  shaft  becomes 
90°,  it  is  a  "quarter-twist"  belt.  To  make  this  clearer,  see 
Fig.  174.  Rotation  is  transmitted  from  A  to  B  by  an  open  belt, 
and  it  is  required  to  turn  the  axis  of  B  out  of  parallelism  with 


s  D' 


FIG.  173. 


FIG.  174. 


that  of  A.  The  direction  of  rotation  is  as  indicated  by  the 
arrows.  Draw  the  line  CD.  If  now  the  line  CD  is  supposed 
to  pass  through  the  center  of  the  belt  at  C  and  D,  it  may 
become  an  axis,  and  the  pulley  B  and  the  part  of  the  belt  FC 
may  be  turned  about  it,  while  the  pulley  A  and  the  part  of  the 
belt  ED  remain  stationary.  During  this  motion  the  center  line 
of  the  part  of  the  belt  CF,  which  is  the  part  that  advances  toward 
the  pulley  B  when  rotation  occurs,  is  always  in  a  plane  perpen- 
dicular to  the  axis  of  the  pulley  B.  The  part  ED,  since  it  has 
not  been  moved,  has  also  its  center  line  in  a  plane  perpendicular 
to  the  axis  of  A.  Therefore  the  pulley  B  may  be  swung  into 
any  angular  position  about  CD  as  an  axis,  and  the  condition 
of  proper  belt  transmission  will  not  be  interfered  with. 

163.  If  the   axes  intersect  the  motion  can  be  transmitted 


BELTS.  235 

between  them  by  belting  only  by  the  use  of  "guide"  or  "idler" 
pulleys.  Let  AB  and  CD,  Fig.  175,  be  intersecting  axes,  and 
let  it  be  required  to  transmit  motion  from  one  to  the  other  by 
means  of  a  belt  running  on  the  pulleys  E  and  F.  Draw  center 
lines  EK  and  FH  through  the  pulleys.  Draw  the  circle,  G, 
of  any  convenient  size,  tangent  to  the  lines  EK  and  FH.  In 
the  axis  of  the  circle,  G,  let  a  shaft  be  placed  on  which  are  two 
pulleys,  their  diameters  being  equal  to  that  of  the  circle,  G. 


FIG.  175. 

These  will  serve  as  guide-pulleys  for  the  upper  and  lower  sides 
of  the  belt,  and  by  means  of  them  the  center  lines  of  the  advanc- 
ing parts  of  both  sides  of  the  belt  will  be  kept  in  planes  perpen- 
dicular to  the  axis  of  the  pulley  toward  which  they  are  advancing, 
the  belts  will  run  properly,  and  the  motion  will  be  transmitted, 
as  required. 

The  analogy  between  gearing  and  belting  for  the  trans- 
mission of  rotary  motion  has  been  mentioned  in  an  earlier 
paragraph.  Spur-gearing  corresponds  to  an  open  or  crossed 
belt  transmitting  motion  between  parallel  shafts.  Bevel-gears 
correspond  to  a  belt  running  on  guide-pulleys  transmitting 
motion  between  intersecting  shafts.  Skew  bevel  and  spiral  gears 
correspond  to  a  "twist"  belt  transmitting  motion  between  shafts 
that  are  neither  parallel  nor  intersecting. 


236  MACHINE  DESIGN. 

164.  Crowning  Pulleys.  —  If  a  flat  belt  is  put  on  a  "crowning  " 
_AO     pulley,  as  in   Fig.   176,  the  tension  on  AB  will  be 

greater  than  on  CD.  The  belt  lying  flat  at  AC  will 
have  its  approaching  portion  bent  out  as  indicated  by 
AE  and  CF,  and  as  rotation  goes  on  the  belt  will 
be  carried  toward  the  high  part  of  the  pulley,  i.e., 
it  will  tend  to  run  in  the  middle  of  the  pulley. 
This  is  the  reason  why  nearly  all  belt  pulleys,  except 
those  on  which  the  belt  has  to  be  shifted  into  different 
positions,  are  turned  "crowning." 

165.  Cone  Pulleys.  —  In  performing  different  operations  on  a 
machine  or  the  same  operations  on  materials  of  different  degrees 
of  hardness,  different  speeds  are  required.     The  simplest  way 
of  obtaining  them  is  by  use  of  cone  pulleys.     One  pulley  has  a 
series  of  steps,  and  the  opposing  pulley  has  a  corresponding 
series  of  steps.     By  shifting  the  belt  from  one  pair  to  another 
the  velocity  ratio  is  changed.     Since  the  same  belt  is  used  on 
all  the  pairs  of  steps,  these  must  be  so  proportioned  that  the 
belt  length  for  all  the  pairs  shall  be  the  same  ;  otherwise  the  belt 
would  be  too  tight  on  some  of  the  steps  and  too  loose  on  others. 
Let  the  case  of  a  crossed  belt  be  first  considered.     The  length 
of  a  crossed  belt  may  be  expressed  by  the  following  formula: 
Let  L=  length  of  the  belt;   d=  distance  between  centers  of  rota- 
tion;   R=  radius  of  the  larger  pulley;    r  =  radius  of  the  smaller 
pulley.     (See  Fig.  177.)     Then 

L  =  2V  d2-  (R  +  r)2  +  (R  +  r)(n  +  2  arc  whose  sine  is 


In  the  case  of  a  crossed  belt,  if  the  size  of  steps  is  changed  so 
that  the  sum  of  their  radii  remains  constant,  the  belt  length 
will  be  constant.  For  in  the  formula  the  only  variables  are  R 
and  r,  and  these  terms  only  appear  in  the  formula  as  R  +  r;  but 
R  +  r  is  by  hypothesis  constant.  Therefore  any  change  that  is 


BELTS. 


237 


made  in  the  variables  R  and  r,  so  long  as  their  sum  is  constant, 
will  not  affect  the  value  of  the  equation,  and  hence  the  belt 
length  will  be  constant.  It  will  now  be  easy  to  design  cone 
pulleys  for  a  crossed  belt.  Suppose  a  pair  of  steps  given  to 
transmit  a  certain  velocity  ratio.  It  is  required  to  find  a  pair 
of  steps  that  will  transmit  some  other  velocity  ratio,  the  length 
of  belt  being  the  same  in  both  cases.  Let  R  and  r  =  radii  of  the 
given  steps;  R'  and  r1  =  radii  of  the  required  steps;  R  +  r  = 
R'+r'=a;  the  velocity  ratio  of  R'  to  r1  =b.  There  are  two 


equations  between  R'  and  r',  R'  +-r'  =b,  and  R'  +  r'=a.     Com- 
bining and  solving,  it  is  found  that  r'  =a  +  (i  +b),  and  R'  =a  —  r'. 
For  an  open  belt  the  formula  for  length,  using  same  symbols 
as  for  crossed  belt,  is 


—  r)  (arc  whose  sine  is  (R  — 


If  R  and  r  are  changed  as  before  (i.e., 
constant),  the  term  R  —  r  would  of  course  not  be  constant,  and 
two  of  the  terms  of  the  equation  would  vary  in  value;  therefore 
the  length  of  the  belt  would  vary.  The  determination  of  cone 
steps  for  open  belts  therefore  becomes  a  more  difficult  matter, 
and  approximate  methods  are  almost  invariably  used. 

166.  Graphical  Method  for  Cone-pulley  Design.—  The  fol- 
lowing graphical  approximate  method  is  due  to  Mr.  C.  A.  Smith, 
and  is  given,  with  full  discussion  of  the  subject,  in  "Transactions 


238 


MACHINE  DESIGN. 


of  the  American  Society  of  Mechanical  Engineers,"  Vol.  X, 
p.  269.  Suppose  first  that  the  diameters  of  a  pair  of  cone  steps 
that  transmit  a  certain  velocity  ratio  are  given,  and  that  the 
diameters  of  another  pair  that  shall  serve  to  transmit  some 
other  velocity  ratio  are  required.  The  distance  between  centers 
of  axes  is  given.  (See  Fig.  178.)  Locate  the  pulley  centers  O 


FIG.  178. 

and  O'  at  the  given  distance  apart;  about  these  centers  draw 
circles  whose  diameters  equal  the  diameters  of  the  given  pair  of 
steps;  draw  a  straight  line  GH  tangent  to  these  circles;  at  J, 
the  middle  point  of  the  line  of  centers,  erect  a  perpendicular, 
and  lay  off  a  distance  JK  equal  to  the  distance  between  centers, 
C,  multiplied  by  the  experimentally  determined  constant  0.314; 
about  the  point  K  so  determined,  draw  a  circular  arc  AB  tan- 
gent to  the  line  GH.  Any  line  drawn  tangent  to  this  arc  will  be 
the  common  tangent  to  a  pair  of  cone  steps  giving  the  same 
belt  length  as  that  of  the  given  pair.  For  example,  suppose 
that  OD  is  the  radius  of  one  step  of  the  required  pair;  about  O, 
with  a  radius  equal  to  OD,  draw  a  circle ;  tangent  to  this  circle 
and  the  arc  AB  draw  a  straight  line  DE;  about  O'  and  tangent 
to  DE  draw  a  circle  \  its  diameter  will  equal  that  of  the  required 
step. 

But  suppose  that,  instead  of  having  one  step  of  the  required 
pair  given,  to  find  the  other  corresponding  as  above,  a  pair  of 


BELTS.  239 

steps  are  required  that  shall  transmit  a  certain  velocity  ratio, 
=  r,  with  the  same  length  of  belt  as  the  given  pair.  Suppose 
OD  and  O'E  to  represent  the  unknown  steps.  The  given  velocity 

ratio    equals   r.      Also,  r=~QTp-      But  from   similar    triangles 

FO 

OD+0'E  =  FO+FOf.     Therefore    r=~;    but    FO=C+x, 

rU 

and  FO'=x.     Therefore  r  = -,  and  x  =  —          Hence  with 

x  r—i 

r  and  C  given,  the  distance  x  may  be  found,  and  a  point  F  located, 
such  that  if  from  F  a  line  be  drawn  tangent  to  AB,  the  cone 
steps  drawn  tangent  to  it  will  give  the  velocity  ratio,  r,  and  a 
belt  length  equal  to  that  of  any  pair  of  cones  determined  by  a 
tangent  to  AB.  The  point  F  often  falls  at  an  inconvenient 
distance.  The  radii  of  the  required  steps  may  then  be  found 
as  follows:  Place  a  straight-edge  tangent  to  the  arc  AB  and 
measure  the  perpendicular  distances  from  it  to  O  and  O'. 
The  straight-edge  may  be  shifted  until  these  distances  bear 
the  required  relation  to  each  other. 

167.  Design  of  Belts. — Fig.  179  represents  two  pulleys 
connected  by  a  belt.  When  no  moment  is  applied  tending  to 
produce  rotation  this  tension  in  the  two  sides  of  the  belt  is 
practically  equal.  Let  T$  represent  this  tension.  If  now 
an  increasing  moment,  represented  by  Rl,  be  applied  to  the 
driver,  its  effect  is  to  increase  the  tension  in  the  lower  side  of 
the  belt  and  to  decrease  the  tension  in  the  upper  side.  With 
the  increase  of  Rl  this  difference  of  tension  increases  till  it  is 
equal  to  P,  the  force  with  which  rotation  is  resisted  at  the  surface 
of  the  pulley.  Then  rotation  begins*  and  continues  as  long  as 
this  equality  continues;  i.e.,  as  long  as  Ti  —  T2=P,  in  which 
TI  =  tension  in  the  driving  side  and  T2=  tension  in  the  slack 

*  While  the  moving  parts  are  being  brought  up  to  speed  the  difference    of 
tension  must  equal  P  plus  force  necessary  to  produce  the  acceleration. 


240 


MACHINE  DESIGN. 


side.    (See  Fig.  180.)    The  tension  in  the  driving  side  is  increased 

T    _L  T 

at  the  expense  of  that  in  the  slack  side.     Therefore  —       —  =  T*. 

2 

rr\ 

To  find  the  value  of  — .     The  increase  in  tension  from  the 

slack  side  to  the  driving  side  is  possible  because  of  the  frictional 
resistance  between  the  belt  and  pulley  surface.     Consider  any 

FIG.  179. 


FIG.  180. 

element  of  the  belt,  ds,  Fig.  181.     It  is  in  equilibrium  under 
the  action  of  the  following  forces: 

T,  the  value  of  the  varying  tension  at  one  end  of  ds; 

T+dT,  the  value  of  the  varying  tension  at  the  other  end  of  ds: 

cds,  the  centrifugal  force; 

pds,  the  pressure  between  the  face  of  the  pulley  and  ds; 

dF  =  [ipds,   the   friction   between    the   element   of   belt   and 
pulley  face. 

Let  the  value  of  cds  be  first  determined. 

Let  A  =  sectional  area  of  belt  in  square  inches ; 
T  =  total  tension  at  any  section  in  pounds; 


BELTS. 


141 


FIG.  181. 

t= allowable  working  tension  in  pounds  per  square  inch 

T 

of  cross-section  of  belt,  so  that  A=—'} 

w  =  weight  of  belt  per  cubic  inch; 

V  =  velocity  of  belt  in  feet  per  minute ; 

v  =  velocity  of  belt  in  feet  per  second; 
R  =  radius  of  pulley  in  feet; 

r  =  radius  of  pulley  in  inches; 

g  =  acceleration  due  to  gravity  =  32. 2  feet  per  second  per 

second ; 
C= centrifugal  force  per  cubic  inch  of  belt  in  pounds; 

C= centrifugal  force  per  linear  inch  of  belt  in  pounds. 


242  MACHINE  DESIGN. 

mass  X  velocity2 

Centrifugal  force  =  — r. ; —        — • 

radius  of  curvature 

TV  V2 

C  =  —  -p-  in  foot-pounds-second  system. 

W    V2       T    W    V2       ~L2lW2    T 

•'•  c^A~g"R=~t'^'^^~gT'7' 

As  an  abbreviation  let  z 


.'.  c  =  T-,    and    cJ5  =  J1—  <&„ 

rrt 

Now,  to  find  the  value  of  -=r: 

J-  2 

Since  dT  is  small  compared  with  T,  no  appreciable  error  is 
introduced  by  considering  pds  as  acting  normally  to  the  chord 
of  ds,  and  we  may  write  the  equation  (sum  of  vert.  comp.  =o, 
in  Fig.  181) 


a  =  total  arc  of  contact  in  degrees. 

6  is  the  total  arc  of  contact  in  radians  =0.01750:. 

dd  is  so  small  that  sin  —  may  be  considered  equal  to  —  in 

radians.  Also  dT  and  dd  both  being  very  small  compared  to 
the  other  quantities,  any  terms  containing  their  product  may  be 
dropped.  Therefore 


.:  pds  =  Tdd-cds  =  Tdd-T- 
But    ds  =  rd6. 


BELTS.  243 

Summing  the  moments  about  o  (Fig.  181)  gives 
T+dT~T+ftpds. 

.'.  dT   = 
dT 


rn 

log.    r  =  Xi- 


and          common  log  ^  =  .  4343^(1  -z)6. 

The  following  equations  are  now  established: 


Tl-T2=P,     .........     (i) 

2T3,     ........     (2) 


log  -^  =  .4343^(1-2)0  ......     (3) 

When  the  belt  speed  is  under  2000  feet  per  minute  the  value 
of  z  is  so  small  that  it  may  be  neglected,  and  equation  (3)  becomes 

T1 

log  ^  =  -4343^- 

The  right-hand  members  of  (i)  and  (3)  can  usually  be  deter- 
mined; hence  the  value  of  T  (the  maximum  stress  in  the  belt) 
may  be  found,  and  proper  proportions  may  be  given  to  the  belt. 
If  W  foot-pounds  per  minute  are  to  be  transmitted,  and  the 
velocity  of  the  rim  of  the  pulley  transmitting  this  power  in  feet 
per  minute  equals  V,  then  the  force  P  equals  the  work  divided 

W 
by  the  velocity,  or  P  =  -y.     The  value  of  a  is  found  from  the 

diameters  of  the  pulleys  and  their  distance  between   centers,. 


•244 


MACHINE  DESIGN. 


and  may  usually  be  estimated  accurately  enough.  The  value 
of  /z,  the  coefficient  of  friction,  varies  with  the  kind  of  belting, 
the  material  and  character  of  surface  of  the  pulley,  and  with 
the  rate  of  slip  of  the  belt  on  the  pulley.  Experiments  made 
at  the  laboratory  of  the  Massachusetts  Institute  of  Technology, 
under  the  direction  of  Professor  Lanza,  indicate  that  for  leather 
belting  running  on  turned  cast-iron  pulleys,  the  rate  of  slip  for 
efficient  driving  is  about  3  to  4  feet  per  minute;  and  also  that 
the  coefficient  of  friction  corresponding  to  this  rate  of  slip  is 
about  0.27.  The  value  0.3  may  be  used.  If  this  value  of  n 
be  used,  the  slip  will  be  kept  within  the  above  limits,  if  the  belt 

be  put  on  with  a  proper  initial  tension,  =  T^  =  —     — ,  and  the 

driving  of  the  belt  so  designed  will  be  satisfactory. 

The  weight  of  leather  belting  on  the  average  per  cubic  inch 
=  w  =  0.035  Ib. 

The  value  of  the  allowable  tension  varies  with  the  quality  of 
the  belt,  the  nature  of  the  splice,  and  other  items.  £  =  300  Ibs. 
per  square  inch  is  a  good  safe  value  for  ordinary  conditions. 
This  is  about  equivalent  to  70  Ibs.  per  inch  of  width  of  single 
belt.  The  following  table  of  values  of  z  has  been  computed 
with  these  values  of  w  and  t.  V  is  the  velocity  of  the  belt  in 
feet  per  minute,  and  v  is  the  velocity  in  feet  per  second. 

TABLE  XIX. — VALUES  OF  z. 


V  ..  . 

l8oO 

2400 

3000 

3600 

4200 

4800 

5400 

6000 

6600 

7200 

7800 

•v.  .  .  . 

30 

40 

5° 

60 

70 

80 

90 

100 

no 

120 

I30 

2.  ... 

-°39 

.070 

.118 

•157 

.214 

.279 

•352 

•435 

.526 

.626 

•735 

It  becomes  evident  that  i  —  z  decreases  rapidly  as  the  velocity 
of  the  belt  increases.  In  fact,  unless  a  greater  value  of  /  than 
300  Ibs.  per  square  inch  is  used  a  leather  belt  weighing  0.035  Ib. 
per  cubic  inch  will  cause  z  to  have  a  value  of  unity  at  about 


BELTS.  245 

9000  feet  per  minute.  This  would  make  the  term  i— z  reduce 
to  zero.  In  other  words  the  belt  can  transmit  no  power  at  this 
speed  because  the  centrifugal  force  is  so  great  that  no  pressure 
exists  between  the  belt  and  the  face  of  the  pulley  and  hence 
there  is  no  friction.* 

168.  Problem. — A  single-acting  pump  has  a  plunger  8  inches 
=0.667  foot  in  diameter,  whose  stroke  has  a  constant  length  of 
10  inches  =0.833  foot.  The  number  of  strokes  per  minute  is  50. 
The  plunger  is  actuated  by  a  crank,  and  the  crank-shaft  is 
connected  by  spur-gears  to  a  pulley-shaft,  the  ratio  of  gears 
being  such  that  the  pulley-shaft  runs  300  revolutions  per  minute. 
The  pulley  which  receives  the  power  from  the  line-shaft  is  18 
inches  in  diameter.  The  pressure  in  the  delivery-pipe  is  100  Ibs. 
per  square  inch.  The  line-shaft  runs  150  revolutions  per 
minute,  and  its  axis  is  at  a  distance  of  12  feet  from  the  axis  of 
the  pulley-shaft. 

Since  the  line-shaft  runs  half  as  fast  as  the  pulley-shaft,  the 
diameter  of  the  pulley  on  the  line-shaft  must  be  twice  as  great 
as  that  on  the  pulley-shaft,  or  36  inches.  The  work  to  be  done 
per  minute,  neglecting  the  friction  in  the  machine,  is  equal  to 
the  number  of  pounds  of  water  pumped  per  minute  multiplied 
by  the  head  in  feet  against  which  it  is  pumped.  The  number 
of  cubic  feet  of  water  per  minute,  neglecting  "slip,"  equals  the 
displacement  of  the  plunger  in  cubic  feet  multiplied  by  the 

0.6672  X- 
number  of  strokes  per  minute  =  —          —  Xo.833X5o  =  i4-55, 

and  therefore  the  number  of  pounds  of  water  pumped  per  minute 
=  14.55X62.4  =  908.  One  foot  vertical  height  or  "head"  of 
water  corresponds  to  a  pressure  of  0.433  lb.  Per  square  inch, 
and  therefore  100  Ibs.  per  square  inch  corresponds  to  a  "head" 

*  The  method  of  computing  belts  in  this  section  is  open  to  minor  theoretical 
objections.  Its  value  lies  in  the  fact  that  it  gives  equations  of  a  simple  form 
which  can  be  applied  easily  and  which  give  results  well  within  the  limits  of 
accuracy  set  by  the  variations  in  the  physical  properties  of  the  belt  materials. 


246  MACHINE  DESIGN. 

of  100-^-0.433  =  231  feet.  The  work  done  pe.  minute  in  pump- 
Ing  the  water  therefore  is  equal  to  908  Ibs. X23i  Let  =209,748 
ft.-lbs.  The  velocity  of  the  rim  of  the  belt-pulley  is  equal  to 
300  XL 5X71=1414  feet  per  minute.*  Therefore  the  force  P  =  T\ 
—  7*2  =  209,748  ft.-lbs.  per  minute -M4I4  feet  per  minute  = 
148  Ibs. 


To    find    a    (see    Fig.     182)     sin/?  =  — y— = -  =  0.0625. 

Therefore  /?  =  3°  3 5';    a  =  i8o°-2/9  =  i8o0-7°  io'  =  i72°  50';    a 
in  TT  measure  =  172!  X 0.0175  =3.025  =  6. 

log  -^  =0.4343  X/*X0  =0.4343X0.3X3.02 5  =0.3941. 

.*.  "^""  =  2.485   f  =  J-  i     _/2  =  i48- 
T2 

Combining  these  equations  T\  is  found  to  be  equal  to  248  Ibs., 
the  maximum  stress  in  the  belt. 

The  cross-sectional  area  of  belt  should  be  equal  to  —  =  — 

t      300 

0.83  square  inch. 

Single-thickness  belting  varies  from  0.2  to  0.25  of  an  inch  in 
thickness,  hence  the  width  called  for  by  our  problem  would  be 

0.83 

-  =  4  inches,  say. 

169.  Problem. — A  sixty -horse-power   dynamo  is  to  run  1500 
revolutions  per  minute  and  has    a  1 5-inch  pulley  on  its  shaft. 

*  At  this  speed  the  simple  form  of  the  belt  formula  may  be  used. 


BELTS.  247 

Power  is  supplied  by  a  line-shaft  running  150  revolutions  per 
minute.     A  suitable  belt  connection  is  to  be  designed. 

The  ratio  of  angular  velocities  of  dynamo-shaft  to  line-shaft 
is  10  to  i;  hence  the  diameter  of  the  pulley  on  the  line-shaft 
would  have  to  be  ten  times  as  great  as  that  of  the  one  on  the 
dynamo,  =12.5  feet,  if  the  connections  were  direct.  This  is 
inadmissible,  and  therefore  the  increase  in  speed  must  be  ob- 
tained by  means  of  an  intermediate  or  counter  shajt.  Suppose 
that  the  diameter  of  the  largest  pulley  that  can  be  used  on  the 
counter-shaft  =48  inches.  Then  the  necessary  speed  of  the 

counter-shaft  =  1  500  X—  5=470,  nearly.     The  ratio  of  diameters 
40 

of  the  required  pulleys  for  connecting  the   line-shaft  and  the 

470 
counter-shaft=—  —  =  3.13.     Suppose  that  a  6o-inch  pulley  can 

be  used  on  the  line-shaft,  then  the    diameter  of   the  required 
pulley  for  the  counter-shaft  will=  —  —  =19  inches,  nearly.    Con- 

sider first  the  belt  to  connect  the  dynamo  to  the  counter-shaft. 
The  work  =  60X33,000  =  1,980,000  ft.-lbs.  per  minute;    the  rim 

of  the  dynamo-pulley  moves  -  X  1500  =  5890  feet  per  minute. 

Therefore  Ti-T2  =  *9  f**    =  336  Ibs.     The  axis  of  the  counter- 

shaft is  10  feet  from  the  axis  of  the  dynamo,  and,  as  before, 
R  —  r    24  —  7.5 

- 


Therefore         /?  =  7°  54'. 

a  =  180°  -2/9  =  164°  12', 
6  =  i64°.2  Xo.oi75  =2.874. 

The  nearest  value  of  V,  in  the  table  for  values  of  z  (p.  244), 
to  5890  is  6000,  and  the  corresponding  value  of  z  is  0.435. 
Hence  1—2  =  0.565. 

rn 

log  -      =  0.4343X0.3X0.565X2.874=0.2111; 


248  MACHINE  DESIGN. 

.:  £-,.6*6. 

-12 


But  Ti-r2  =  3361bs. 

/.  0.626X2  =  336  Ibs. 

•"•  ^2  =  537  Ibs. 

.-.  7^1=873  Ibs. 

The  cross-sectional  area  of  the  belt=  -  =  2.9  square  inches. 

A  double  belt  is  about  f  inch  thick.     Our  problem,  then, 
calls  for  a  double  belt  1X2.9=8  inches,  say,  wide. 

170.  Variation   of   Driving    Capacity.  —  From    equation    (3), 

T* 

p.  231,  it  follows  that  the  ratio  of  tensions,  -=r  ,  when  the  belt 

slips  at  a  certain  allowable  rate  (i.e.,  when  /z  is  constant),  de- 
pends only  upon  a.  The  velocity  of  the  belt  also  remains 
constant.  This  ratio,  therefore,  is  independent  of  the  initial 

T* 

tension,   T3;    hence  "taking  up"  a  belt  does  not  change   —  . 

T2 

The  difference  of  tension,  T\  —  T<t=P,  is,  however,  dependent 
on  TV  Because  p,  the  normal  pressure  between  belt  and 
pulley,  varies  directly  as  T3,  then,  since  dF  =  jipds  =  dT,  it 
follows  that  dT  varies  with  T3,  and  hence 


varies  with  TV     This  is  equivalent  to  saying  that  "taking  up  " 
a  belt  increases  its  driving  capacity. 

This  result  is  modified  because  another  variable  enters  the 
problem.  If  jT3  is  changed,  the  amount  of  slipping  changes, 
and  the  coefficient  of  friction  varies  directly  with  the  amount 
of  slipping.  Therefore  an  increase  of  7^  would  increase  p 


BELTS.  249 

and  decrease  //  in  the  expression  npds=dT,  and  the  converse 
is  also  true.     This  is  probably  of  no  practical  importance. 

The  value  of  P  may  also  be  increased  by  increasing  either  p, 
the  coefficient  of  friction,  or  d,  the  arc  of  contact,  since  increase 

/y-i 

of    either    increases    the    ratio  -=r,    and     therefore    increases 

12 

Tl-T2=P. 

Increasing  T3  decreases  the  life  of  the  belt.  It  also  in- 
creases the  pressure  on  the  bearings  in  which  the  pulley-shaft 
runs,  and  therefore  increases  frictional  resistance;  hence  a 
greater  amount  of  the  energy  supplied  is  converted  into  heat  and 
lost  to  any  useful  purpose.  But  if  T3  is  kept  constant,  and  ft 
or  0  is  increased,  the  driving  power  is  increased  without  increase 
of  pressure  in  the  bearings,  because  this  pressure  =  2  Ts  remains 
constant.  When  possible,  therefore,  it  is  preferable  to  increase 
P  by  increase  of  p  or  6,  rather  than  by  increase  of  TV 

Application  of  belt-dressing  may  serve  sometimes  to  in- 
crease fi. 

If,  as  in  Fig.  179,  the  arrangement  is  such  that  the  upper 
side  of  the  belt  is  the  slack  side,  the  "sag  "  of  the  belt  tends  to 

T 
increase  the  arc  of  contact,  and  therefore  to  increase  -^r.      If 

J-  2 

the  lower  side  is  the  slack  side,  the  belt  sags  away  from  the  pulleys 

rr\ 

and  6  and  -^r  are  decreased. 
1  2 


FIG    183. 

An  idler-pulley,  C,  may  be  used,  as  in  Fig.  183.     It  is  pressed 
against  the  belt  by  some  means.     Its  purpose  may  be  to  increase 

T"  -\-  T" 
P  by  increasing  the  tension,  T3  =—    — -.      In  this  case  friction 


250  MACHINE   DESIGN. 

in  the  bearings  is  increased,  and  this  method  should  be  avoided. 
Or  it  may  be  used  on  a  slack  belt  to  increase  the  angle  of  con- 

T 
tact,  a,  the  ratio  -=r»  and  therefore  P,  the  driving  force.     In 

1  2 

T  4-  7~" 
this  case  the   value  of    T3,  =—      — ,  may  be  made  as  small  a 

value  as  is  consistent  with  driving,  and  hence  the  journal  friction 
may  be  small. 

Tighteners  are  sometimes  used  with  slack  belts  for  dis- 
engaging gear,  the  driving-pulley  being  vertically  below  the 
follower. 

In  the  use  of  any  device  to  increase  /z  and  a,  it  should  be 
remembered  that  7i  is  thereby  increased,  and  may  become 
greater  than  the  value  for  which  the  belt  was  designed.  This 
may  result  in  injury  to  the  belt. 

In  Fig.  184,  the  smaller  pulley,  A,  is  above  the  larger  one, 
B.  A  has  a  smaller  arc  of  contact,  and  hence  the  belt  would 


FIG.  184.  FIG.  185. 

slip  upon  it  sooner  than  upon  B.  The  weight  of  the  belt,  how- 
ever, tends  to  increase  the  pressure  between  the  belt  and  A, 
and  to  decrease  the  pressure  between  the  belt  and  B.  The 
driving  capacity  of  A  is  thereby  increased,  while  that  of  B  is 
diminished;  or,  in  other  words,  the  weight  of  the  belt  tends  to 
equalize  the  inequality  of  driving  power.  If  the  larger  pulley 
had  been  above,  there  would  have  been  a  tendency  for  the 
belt  weight  to  increase  the  inequality  of  driving  capacity  of  the 


BELTS.  251 

pulleys.     The  conclusion  from  this,  as  to  arrangement  of  pulleys, 
is  obvious. 

171.  Proper  Size   of  Pulleys. — A  belt  resists  a  force  which 
tends  to  bend  it.     Work  must  be  done,  therefore,  in  bending  a 
belt  around  a  pulley.     The  more  it  is  bent  the  more  work  is 
required  and  the  more  rapidly  the  belt  is  worn  out.     Suppose 
A B,  Fig.  185,  to  represent  a  belt  which  moves  from  A  toward 
B.      If  it  runs  upon  C  it    must  be  bent  more  than  if  it  runs 
upon  D.     The  work  done  in  bending  the  belt  is  converted  into 
useless  heat  by  the  friction  between  the  belt  fibers.    It  is  desirable, 
therefore,  to  do  as  little  bending  as  possible.    This  is  one  reason 
why  large  pulleys  in  general  are  more  efficient  than  little  ones. 
The  resistance  to  bending  increases  with  the  thickness  of  the 
belt,  and  hence  double  belts  should  not  be  used  on  small  pulleys 
if  it  can  be  avoided. 

Double        belts  may  be  used  on  pulleys  12"  and  over. 
Triple  "        "     "      "     "        "      20"    " 

Quadruple     "       "     "     "     "        "      30"    "       " 

172.  Distance  Desirable  between   Shafts. — In  the  design   of 
belting  care  should  be  taken  not  to  make  the  distance  between 
the  shafts  carrying  the  pulleys  too  small,  especially  if  there  is  the 
possibility  of  sudden  changes  of  load.     Belts  have  some  elasticity, 
and  the  total  yielding  under  any  given  stress  is  proportional  to 
the  length,  the  area  of  cross-section  being  the  same.     There- 
fore a  long  belt  becomes  a  yielding  part,  or  spring,  and  its  yielding 
may  reduce  the  stress  due  to  a  suddenly  applied  load  to  a  safe 
value ;  whereas  in  the  case  of  a  short  belt,  with  other  conditions 
exactly  the  same,  the  stress  due  to  much  less  yielding  might  be 
sufficient  to  rupture  or  weaken  the  joint. 

173.  Rope-drives. — The  formulae  which  have  been  derived 
for  belts  also  apply  to  rope-drives.      For  good  durability  the 
allowable  tension  in  .a  rope-drive  should    be  about  20od2  Ibs. 
where  d  is  the  diameter  of  the  rope  in  inches.     Experiments 


252 


MACHINE  DESIGN. 


vary  greatly  in  the  value  of  the  coefficient  of  friction  for  a  well- 
lubricated  rope  on  a  flat-surfaced  smooth  metal  pulley.  It  may 
be  taken  equal  to  0.12.*  But  ropes  are  not  commonly  used 
on  flat  pulleys;  instead  of  this  they  run  in  grooves  on  the  faces 
of  sheave  wheels,  and  substitution  must  be  made  for  fi  in  the 

angle  of  groove 
formula  (3),  not  0.12  but  o.i2Xcosec —     — . 

The  following  table  gives  the  values  of  fi  for  different  angles 

of  grooves. 

TABLE  XX. 


Angle  of  groove  in  degrees  .  .  . 

3° 

35 

40 

45 

5° 
o  28 

55 

60 

Taking  the  weight  of  rope  per  linear  inch  =o.O268d2  Ibs.,  and 
the  allowable  tension  =20od2  Ibs.,  and  solving  for  z  at  various 
speeds,  gives  the  following  results: 


TABLE  XXI. 


V.  .. 

1—2. 

IOOO 

0.98 

2000 
0.94 

2500 
0.91 

3000 
0.87 

3500 
0.83 

4000 
0.78 

4500 
0.72 

5000 
0.65 

5500 
0.58 

6OOO 
0.50 

6500 
0.41 

7OOO 
0.32 

7500 
0.22 

8000 
O.II 

8500 
0.00 

V  is  the  velocity  of  the  rope  in  feet  per  minute. 
For  convenience  the  following  table  is  given   showing  the 
corresponding  values  of  angles  in  degrees  and  circular  measure: 


TABLE  XXII. 


a.  .. 

e.  .. 

i°5 
i.83 

120 
2.09 

135 

2-35 

15° 
2.62 

165 
2.88 

180 
3-14 

195 
3-43 

210 

3-66 

240 
4.19 

It  will  be  remembered  that  #=0.01750:. 

The  diameter  of  the  sheave  wheel  is  properly  calculated  from 
the  point  of  tangency  of  the  rope  to  the  groove  (A- A,  Fig.  186) 
and  not  from  the  middle  of  the  rope  O.  The  diameter  of  the 
sheave  wheel  should  not  be  too  small  or  the  rope  will  wear  out 


*  "Rope -driving,"  by  J.  J.  Flather.   New  York:  Wiley  &  Sons. 


BELTS. 


253 


very  rapidly.     The  following  table  gives  the  minimum  values, 
D  being  the  diameter  of  the  wheel  and  d  that  of  the  rope : 
TABLE  XXIII. 


d  
D  

i" 

24 

i" 
36 

ii" 
48 

ii" 
60 

if" 

72 

2" 
84 

It  is  still  better  to  make  D=$od. 

Two  systems  of  rope-driving  are  in  use,  the 
English  and  the  American.  In  the  former  a  number 
of  ropes  are  used  side  by  side.  In  the  latter  a 
single  continuous  rope  is  used  with  a  guide  and 
tightener.  As  long  as  all  the  grooves  in  each 
sheave  are  alike,  each  rope  will  tend  to  carry  its 
proportionate  share  of  the  load  in  the  English 
system,  provided  all  the  ropes  had  the  same 
original  tension,  and  have  stretched  the  same 
amount. 

For  the  American  system  the  grooves  in  the 
larger  sheave  should  have  a  greater  angle  than 
those  in  the  smaller  sheave  or  the  load  will  be 
unequally  divided  among  the  various  wraps  of  the 
rope.* 

The  tension  in  each  wrap  of  the  rope  will  be  the 
same,  when  running,  if  the  friction  on  each  sheave 
wheel  is  the  same.  The  friction  on  each  pulley 
will  be  the  same  if  the  products  of  the  arcs 
of  contact  by  the  respective  coefficients  are 
equal. 

Let  ju  =  coefficient  of  larger  sheave  =0.12  cosec  - ;       FIG.  186. 

/•  =  angle  of  groove  of  larger  sheave; 
a=arc  of  contact  of  larger  sheave; 


*  See  further  Proc.  Am.  Soc    C.  E.,  Vol.  XXIII. 
Rope -driving. 


Mr.  Spencer  Miller  on 


254 


MACHINE  DESIGN. 


ft'  =  coefficient  of  smaller  sheave  =0.12  cosec  — ; 
•f  =  angle  of  groove  of  smaller  sheave; 
a!  =arc  of  contact  of  smaller  sheave. 
Then  fia.  should  =//a'. 

r  fa? 

.'.  cosec  -  =  cosec  —  X— . 
2  2     a 

The  following  table  gives  the  proper  values  for  equal  ac 
hesion : 

TABLE  XXIV. — ANGLE  OF  GROOVE  FOR  EQUAL  ADHESION. 


Arc  of  contact  on  small  pulley     a' 

0.9 

0.8 

0-75 

0.7 

0.65 

0.6 

Arc  of  contact  on  large  pulley     a  '  ' 

Angle  of  groove  in  large  pulley  when 

groove  in  small  pulley  =  35°  

40° 

44° 

47° 

5i° 

55° 

60° 

Angle  of  groove  in  large  pulley  when 

groove  in  small  pulley  =  40°  

45° 

S0° 

54° 

S8° 

64° 

70° 

Angle  of  groove  in  large  pulley  when 

groove  in  small  pulley  =  45°  

50° 

55° 

60° 

66° 

72° 

80° 

The  angle  of  groove  on  the  smaller  sheave  wheel  is  generally 
made  45°.  Assuming  this  an  angle  of  contact  of  165°,  and  a  i 
allowable  stress  =  2Ood2,  the  following  table  has  been  com- 
puted for  the  horse-power  transmitted  by  each  wrap  of  the  rope : 


TABLE  XXV 


Velocity 
of  Rope 
in  Feet 


Diameter  of  Rope  in  Inches 


M&, 

• 

> 

i 

<* 

•» 

•« 

2 

1000 

1.24 

2.25 

3-57 

5-59 

8.02 

10.85 

14.20 

2000 

2.70 

6.84 

10.68 

15-39 

20.93 

27.36 

2500 

3-30 

4.71 

8.38 

13.10 

18.86 

25  .  66 

33  •  54 

3000 

3.83 

5-46 

9.80 

15-39 

21.87 

29-74 

38.88 

3500 

4-30 

6.23 

11.09 

17-33 

24-94 

34-03 

44  •  35 

4000 

4-74 

6.83 

12.15 

18.98 

27-33 

37-17 

48.59 

4500 

5.01 

7.24 

12.89 

20.15 

29.00 

39-45 

5^-57 

5000 

5-20 

7-47 

13.29 

20.76 

29.89 

40.65 

53  »5 

55°0 

5-29 

7.60 

13-53 

21.14 

30-43 

41-39 

54-11 

6000 

5.08 

7-32 

13.10 

20.36 

29-32 

39-77 

52.12 

6500 

4-74 

6.83 

12.13 

19.00 

27-34 

37-21 

48.63 

7000 

4.12 

5-93 

10.54 

16.47 

23-72 

32.26 

42.18 

7500 

3-25 

4.67 

8.32 

13.00 

18.72 

25-42 

33-23 

BELTS.  255 

For  durability  a  few  turns  of  a  larger  rope  are  preferable  to 
more  turns  of  a  smaller  rope. 

The  most  economical  speed,  taking  first  cost  and  relative 
wear  into  consideration,  is  about  4500  feet  per  minute. 


CHAPTER  XVI. 

FLY-WHEELS. 

174.  Theory  of  Fly-wheel.  —  Often  in  machines  there  is 
capacity  for  uniform  effort,  but  the  resistance  fluctuates.  In 
other  cases  a  fluctuating  effort  is  applied  to  overcome  a  uniform 
resistance,  and  yet  in  both  cases  a  more  or  less  uniform  rate 
of  motion  must  be  maintained.  When  this  occurs,  as  has  been 
explained,*  a  moving  body  of  considerable  weight  is  interposed 
between  effort  and  resistance,  which,  because  of  its  weight, 
absorbs  and  stores  up  energy  with  increase  of  velocity  when 
the  effort  is  in  excess,  and  gives  it  out  with  decrease  of  velocity 
when  the  resistance  is  in  excess.  This  moving  body  is  usually 
a  rotating  body  called  a  fly-wheel. 

To  fulfill  its  office  a  fly-wheel  must  have  a  variation  of 
velocity,  because  it  is  by  reason  of  this  variation  that  it  is  able 
to  store  and  give  out  energy.  The  kinetic  energy,  E,  of  a 
body  whose  weight  is  W,  moving  with  a  velocity  v,  is  expressed 
by  the  equation 


To  change  E,  with  W  constant,  v  must  vary.  The  allowable 
variation  of  velocity  depends  upon  the  work  to  be  accomplished. 
Thus  the  variation  in  an  engine  running  electric  lights  or  spin- 
ning-machinery should  be  very  small,  probably  not  greater 

*  See  §43. 

256 


FLY-WHEELS.  257 

than  a  half  of  one  per  cent,  while  a  pump  or  a  punching- 
machine  may  have  a  much  greater  variation  without  interfering 
with  the  desired  result.  If  the  maximum  velocity,  v\,  of  the 
fly-wheel  rim  and  the  allowable  variation  are  known,  the  mini- 
mum velocity,  V2,  becomes  known;  and  the  energy  that  can  be 
stored  and  given  out  with  the  allowable  change  of  velocity  is 
equal  to  the  difference  of  kinetic  energy  at  the  two  velocities. 

Wvf     Wv2*     W. 


. 

2g  2g  2g 

175.  The  general  method  for  fly-wheel  design  is  as  follows: 
Find  the  maximum  energy  due  to  excess  or  deficiency  of  effort 
during  a  cycle  of  action,  =E.  Use  the  foot-pound-second 
system  of  units.  Assume  a  convenient  mean  diameter  of  fly- 
wheel rim.  From  this  and  the  given  maximum  rotative  speed 
of  the  fly-wheel  shaft  find  v\.  Solve  the  above  equation  for 
W  thus: 


Substitute  the  values  of  E,  v\,  v2,  and  #  =  32.2  feet  per  second,2 
whence  W  becomes  known,  =  weight  of  fly-wheel  rim.  The 
weight  of  rim  only  will  be  considered;  the  other  parts  of  the 
wheel,  being  nearer  the  axis,  have  less  velocity  and  less  capacity 
per  pound  for  storing  energy.  Their  effect  is  to  reduce  slightly 
the  allowable  variation  of  velocity.* 

176.  Problem.  —  In    a   punching-machine  the  belt  is  capable 
of  applying  a  uniform  torsional  effort  to  the  shaft;    but  most 


*  Numerical  examples  taken  from  ordinary  medium-sized  steam-engine  fly- 
wheels show  that  while  the  combined  weight  of  arms  and  hub  equals  about  one 
third  of  the  total  weight  of  the  wheel,  the  energy  stored  in  them  for  a  given  varia- 
tion of  velocity  is  only  about  10  per  cent  of  that  stored  in  the  rim  for  the  same 
variation. 


2 58  MACHINE  DESIGN. 

of  the  time  it  is  only  required  to  drive  the  moving  parts  of  the 
machine  against  frictional  resistance.  At  intervals,  however^ 
the  punch  must  be  forced  through  metal  which  offers  shearing 
resistance  to  its  action.  Either  the  belt  or  fly-wheel,  or  the  two 
combined,  must  be  capable  of  overcoming  this  resistance.  A 
punch  makes  30  ntrokes  per  minute,  and  enters  the  die  |  inch.  It 
is  required  to  punch  f-inch  holes  in  steel  plates  ^  inch  thick. 
The  shearing  strength  of  the  steel  is  about  50,000  Ibs.  per  square 
inch.  When  the  punch  just  touches  the  plate  the  surface  which 
offers  shearing  resistance  to  its  action  equals  the  surface  of  the 
hole  which  results  from  the  punching,  =ndt,  in  which  d=  diam- 
eter of  hole  or  punch,  t  =  thickness  of  plate.  The  maximum 
shearing  resistance,  therefore,  equals  7^X^X50,000  =  58,900  Ibs. 
As  the  punch  advances  through  the  plate  the  resistance  decreases 
because  the  surface  in  shear  decreases,  and  when  the  punch 
just  passes  through  the  plate  the  resistance  becomes  zero.  If 
the  change  of  resistance  be  assumed  uniform  (which  would 
probably  be  approximately  true)  the  mean  resistance  to  punching 
would  equal  the  maximum  resistance  +  minimum  resistance,  -^2, 

—  =29,450.     The  radius  of  the  crank  which  actuates, 

the  punch  =  2  in.  In  Fig.  187  the  circle  represents  the  path  of 
the  crank-pin  center.  Its  vertical  diameter  then  represents  the 
travel  of  the  punch.  If  the  actuating  mechanism  be  a  slotted 
cross-head,  as  is  usual,  it  is  a  case  of  harmonic  motion,  and  it  may 
be  assumed  that  while  the  punch  travels  vertically  from  A  to  B, 
the  crank-pin  center  travels  in  the  semicircle  ACB.  Let  BD 
and  DE  each  =  |  inch.  Then  when  the  punch  reaches  E  it 
just  touches  the  plate  to  be  punched,  which  is  £  inch  thick,  and 
when  it  reaches  D  it  has  just  passed  through  the  plate.  Draw  the 
horizontal  lines  EF  and  DG  and  the  radial  lines  OG  and  OF. 
Then,  while  the  punch  passes  through  the  plate,  the  crank-pin 
center  moves  from  F  to  G,  or  through  an  angle  (in  this  case) 


FLY-WHEELS. 


259 


of  19°.  Therefore  the  crank-shaft  A,  Fig.  188,  and  attached 
gear  rotate  through  19°  during  the  action  of  the  punch.  The 
ratio  of  angular  velocity  of  the  pinion  and  the  gear  =  the  inverse 

ratio  of  pitch  diameters  = —  =  5.     Hence  the  shaft  B  rotates 

through  an  angle  =  i9°X5  =95°  during  the  action  of  the  punch. 
If  there  were  no  fly-wheel  the  belt  would  need  to  be  designed 
to  overcome  the  maximum  resistance;  i.e.,  the  resistance  at  the 
instant  when  the  punch  is  just  beginning  to  act.  This  would 


m 

1 

t 

I 
\ 

i  i 

'II 

Crank                      A 
B 

y 

i 

ill 

I 

t  ? 

rf 

FIG.  187. 


FIG.  188. 


give  for  this  case  a  double  belt  about  20  inches  wide.  The  need 
for  a  fly-wheel  is  therefore  apparent.  Assume  that  the  fly- 
wheel may  be  conveniently  36  inches  mean  diameter,  and  that 
a  single  belt  5  inches  wide  is  to  be  used.  The  allowable  maxi- 
mum tension  is  then  =5Xallowable  tension  per  inch  of  width 
of  single  belting  =  5  X 70  =  350  Ibs.  =  T\. 

Since  the  pulley-shaft  makes  150  revolutions  per  minute  and 
the  diameter  of  the  pulley  is  2  feet,  the  velocity  of  the  belt  = 
150X2X^  =  942  feet  per  minute.  At  this  slow  speed  the  simple 

rr\ 

form  of  the  belt  formula  may  be  used,  i.e.,  log  -^r=  0.4343^0. 

i  2 

Assume  an  angle  of  contact  of  180°.     Then 


260  MACHINE  DESIGN. 

0  =  3.1416, 


T 

log  jr  =  0.4081; 

T" 
and  .-.  ^=2.56. 


TI  —  TZ  =  213.3  Ibs.  =the  driving  force  at  the  surface  of  the  pulley. 
Assume  that  the  frictional  resistance  of  the  machine  is 
equivalent  to  25  Ibs.  applied  at  the  pulley-rim.  Then  the  belt 
can  exert  213.3  —  25=188.3  Ibs.  =P,  to  accelerate  the  fly-wheel 
or  to  do  the  work  of  punching.  Assume  variation  of  velocity  =  10 
percent.  The  work  of  punching  =  the  mean  resistance  offered 
to  the  punch  multiplied  by  the  space  through  which  the  punch 

acts,    =  —  —  Xo.5"  =  14725  in.  -Ibs.  =1225  ft.  -Ibs.     The  pulley- 

shaft  moves  during  the  punching  through  95°,  and  the  driv- 
ing tension  of  the  belt,  =P  =  188.3  Ibs.,  does  work=PXspace 

moved  through  during  the  punching  =  188.3  Ibs.  XTT(/-~-  =  188.3 

lbs.X?rX2    ft.  X-^r-  =  312   ft.-lbs.     The   work  left   for  the   fly- 

wheel to  give  out  with  a  reduction  of  velocity  of  10  per  cent 
=  1225—312=913  ft.-lbs.  Let  vi  =  maximum  velocity  of  fly- 
wheel rim;  v2  =  minimum  velocity  of  fly-wheel  rim;  W  =  weight 
of  the  fly-wheel  rim.  The  energy  it  is  capable  of  giving  out, 

W(vl2-v22) 
while  its  velocity  is  reduced  from  v\  to  v2,   =  --  -  -  ,  an:l 

the  value  of  W  must  be  such  that  this  energy  given  out  shall 
equal  913  ft.-lbs.  Hence  the  following  equation  may  be  written: 


FLY-WHEELS.  26r 

Therefore  W  =  g***^**. 

The  punch-shaft  makes  30  revolutions  per  minute  and  the  pulley- 
shaft    30X5  =  150=^   revolutions   per  minute.      Hence   "V\    in 

feet  per  second=—^ — ,  D  be  ing  fly  -wheel  diameter  in  feet  =3  feet. 


Hence  ^^      =555  Ibs. 

To  proportion  the  rim:    A  cubic  inch  of  cast  iron  weighs 

0.26  lb.;    hence  there  must  be  J^^  =  2i35  cu.  ins.     The  cubic 

0.20 

contents  of  the  rim=mean  diameter  XTTX its  cross-sectional  area 
A  =2135  cu.  ins.;  hence 

213^ 

^=:^b  =  l8-45  sq.  ins. 


If  the  cross-section  were  made  square  its  side  would  =v/ 18.45 

=  4-3- 

177.  Pump  Fly-wheel. — The  belt  for  the  pump,  p.  245,  is 
designed  for  the  average  work.  A  fly-wheel  is  necessary  to 
adapt  the  varying  resistance  to  the  capacity  of  the  belt.  The 
rate  of  doing  work  on  the  return  stroke  (supposing  no  resistance 
due  to  suction)  is  only  equal  to  the  frictional  resistance  of  the 
machine.  During  the  working  stroke  the  rate  of  doing  work 
varies  because  the  velocity  of  the  plunger  varies,  although  the 
pressure  is  constant.  The  rate  of  doing  work  is  a  maximum 
when  the  velocity  of  the  plunger  is  greatest.  In  Fig.  189,  A 
is  the  velocity  diagram,  B  is  the  force  diagram,  C  is  the 


•262  MACHINE  DESIGN. 

tangential  diagram  drawn  as  indicated  on  pp.  68-70.  The 
belt,  4  inches  wide,  is  capable  of  applying  a  tangential  force 
of  148  Ibs.  to  the  i8-inch  pulley-rim.  The  velocity  of  the  pulley- 
rim  =  711.5X300  =  1414'.  The  velocity  of  the  crank-pin  axis 
=  71X0.833X50  =  130. 9'.  Therefore  the  force  of  148  Ibs.  at 


the  pulley- rim  corresponds  to  a  force 


Ibs. 


130.9 

applied  tangentially  at  the  crank-pin  axis.      This  may  be  plotted 
as  an  ordinate  upon  the  tangential   diagram  C,  from  the  base 


FIG.  189. 


line  XXij  using  the  same  force  scale.  Through  the  upper 
extremity  of  this  ordinate  draw  the  horizontal  line  DE.  The 
area  between  DE  and  XXi  represents  the  work  the  belt  is 
capable  of  doing  during  the  working  stroke.  During  the  return 
stroke  it  is  capable  of  doing  the  same  amount  of  work.  But 
this  work  must  now  be  absorbed  in  accelerating  the  fly-wheel. 
Suppose  the  plunger  to  be  moving  in  the  direction  shown  by 
the  arrow.  From  E  to  F  the  effort  is  in  excess  and  the  fly-wheel 
is  storing  energy.  From  jP  to  G  the  resistance  is  in  excess  and 
the  fly-wheel  is  giving  out  energy.  The  work  the  fly-wheel  must 
be  capable  of  giving  out  with  the  allowable  reduction  of  velocity 
is  that  represented  by  the  area  under  the  curve  above  the  lire 
FG.  From  G  to  D,  and  during  the  entire  return  stroke,  the 
belt  is  doing  work  to  accelerate  the  fly-wheel.  This  work 


PLY-WHEELS.  263 

becomes  stored  kinetic  energy  in  the  fly-wheel.     Obviously  the 
following  equation  of  areas  may  be  written  : 


XiEF+XGD  +  XHKXi  =  GMF. 

The  left-hand  member  of  this  equation  represents  the  work 
done  by  the  belt  in  accelerating  the  fly-wheel;  the  right-hand 
member  represents  the  work  given  out  by  the  fly-wheel  to  help 
the  belt. 

The  work  in  foot-pounds  represented  by  the  area  GMF 
may  be  equated  with  the  difference  of  kinetic  energy  of  the 
fly-wheel  at  maximum  and  minimum  velocities.  To  find  the 
value  of  this  work:  One  inch  of  ordinate  on  the  force  diagram 
represents  8520  Ibs.;  one  inch  of  abscissa  represents  0.449  foot. 
Therefore  one  square  inch  of  area  represents  8520  Ibs.  X  0.449' 
=  3825.48  ft.  -Ibs.  The  area  GMF  =0.4  sq.  ins.  Therefore 
the  work  =  382  5.  48X0.4  =  1530  ft.  -Ibs.  =E.  The  difference  of 

W 

kinetic  energy  =—  (v}i2—v22)  =1530;    W  equals  the  weight  of 

the  fly-wheel  rim.     Hence 

1530X32.2X2 

o  o  • 

Vi2-V22 

Assume  the  mean  fly-wheel  diameter  =  2.  5  feet.  It  will  be 
keyed  to  the  pulley-shaft,  and  will  run  300  revolutions  per 
minute,  =  5  revolutions  per  second.  The  maximum  velocity 
of  fly-wheel  rim  =77X2.5X5  =39.27  =i>i.  Assume  an  allowable 
variation  of  velocity,  =5  per  cent.  Then  ^2  =  37-27X0.95  = 
37.3;  vi2  =  1542.3;  1/2  =  1391.  3;  Vi2-v#=i$i.  Hence 


1530X32.2X2^ 


There  must  be  651  s-o.26  cu.  in.  in  the  rim,  =2504.     The  mean 
circumference  =30"  XTT=  94.2".     Hence  the  cross-sectional  area 


264  MACHINE  DESIGN. 

of    rim  =2504-^94.2  =26.6  sq.    ins.      The   rim   may   be    made 
4.5"  X  6". 

The  frictional  resistance  of  the  machine  is  neglected.  It 
might  have  been  estimated  and  introduced  into  the  problem  as  a 
constant  resistance. 

178.  Steam-engine  Fly-wheel. — From   given   data  draw  the 
indicator-card  as  modified  by  the  acceleration  of  reciprocating 
parts.     See  page  67  and  Fig.  46.      From  this  and  the  velocity 
diagram   construct   the   diagram   of   tangential    driving   force, 
Fig.  47.    Measure  the  area  of  this  diagram  and  draw  the  equiva- 
lent rectangle  on  the  same  base.     This  rectangle  represents  the 
energy  of  the  uniform  resistance  during  one  stroke;    while  the 
tangential  diagram  represents  the  work  done  by  the  steam  upon 
the    crank-pin.     The    area    of   the    tangential   diagram    which 
extends  above  the  rectangle  represents  the  work  to  be  absorbed 
by  the  fly-wheel  with  the  allowable  variation  of  velocity.*     Find 
the  value  of  this  in  foot-pounds,  and  equate  it  to  the  expression 
for  difference   of   kinetic   energy  at  maximum   and   minimum 
velocity.     Solve  for  W,  the  weight  of  fly-wheel. 

179.  Stresses  in   Fly-wheel  Rims. — Mathematical  analyses 
of  the  stresses  in  fly-wheel  rims  are  unsatisfactory.     In  the  first 
place,  in  order  to  get  solutions  of  reasonable  simplicity  it   is 
customary   to   make    assumptions   which   are    contrary   to   the 
actual   conditions;    and    in    the   second   place,   no   satisfactory 
data  exist  concerning  the  strength  of  cast  iron  in  such  heavy 
sections  as  are  used  in  large  engine  fly-wheels.     An  examination 
of  the  nature  of  the  stresses,  however,  will  indicate  the  points 
to  be  looked  out  for  in  design. 

Considering  a  ring  of  hollow  cylindrical  form,  comparatively 

*  For  compound  engines  and  for  varying  resistances  the  diagrams  should  be 
constructed  for  the  complete  cycle.  For  full  treatment  of  the  problem  of  fly 
wheels  for  engines  driving  alternators  the  reader  is  referred  to  the  Trans.  A.  S. 
M.  E.,  Vol.  XXII,  p.  955,  and  Vol.  XXIV,  p.  98. 


FLY-WHEELS.  265 

thin  radially,  it  can  be  shown  that,  when  it  is  rotated  about  its 
axis,  tension  is  set  up  in  the  ring  proportional  to  the  weight  of 
the  material  used  and  the  square  of  the  linear  velocity.  This 
tension  is  due  solely  to  the  action  of  "centrifugal  force"  and  is 
termed  "centrifugal  tension." 

Consider  the  half- ring  shown  in  Fig.  190: 
^=the  velocity  of  the  rim  in  feet  per  second. 
c  =  "centrifugal  force"  per  foot  of  rim; 
R=  radius  in  feet; 
A  =area  of  rim  in  square  inches; 
P=  total  tension  in  rim  in  pounds; 
ft  =unit  tensile  stress  in  rim  in  pounds; 
w  =  weight    of  material  as  represented  by  a  piece  i  inch 

square  and  i  foot  long; 
g  =32.2  feet  per  second  per  second; 

2.P=sum.  of  horizontal  components  of  all  the  small  centri- 
fugal forces  cds; 

Each  horizontal  component  =  cds  cos  6,  which  may  be  written 
cR  cos  0  dd,  because  ds=Rdd. 


2P= 


:.  P=cR. 
Mv2     Wv2 


But 


g  ' 

W  being  the  weight  of  one  linear  foot  of  rim 


266  MACHINE  DESIGN. 

Also,  P=M; 


For  cast  iron,  putting  ft  =20,000  Ibs.  the  ultimate  strength, 
and  w  =0.26X12  Ibs.,  it  follows  that  ^=454  feet  per  second. 
In  other  words  a  cast-iron  ring  will  burst  at  a  speed  of  454  feet 
per  second.  Furthermore,  an  examination  of  the  formula 
shows  that  for  a  ring  this  bursting  velocity  depends  not  at  all 
on  the  size  or  shape  of  the  cross-section  but  only  on  the  material 
used  as  represented  by  ft  and  w. 

This  centrifugal  tension  causes  a  corresponding  elongation 
of  the  material  and  therefore  an  increase  in  the  radius  of  the 
ring.  A  free  ring  of  whatever  cross-section  can  and  does  take 
the  new  radius  and  the  tension  on  all  sections  =/<  pounds  per 
square  inch. 

With  the  introduction  of  rigidly  fastened  arms  a  number  of 
new  and  vital  elements  enter  into  the  problem.  An  arm  of  the 
same  original  length  as  the  original  radius  of  the  rim  when 
rotated  about  an  axis  perpendicular  to  its  inner  end  will  also 
suffer  an  elongation  due  to  centrifugal  action.  The  amount  of  this 
radial  elongation  will  vary  with  the  form  of  the  arm,  but  in  no 
practical  case  will  it  amount  to  as  much  as  one  third  of  the 
radial  increase  of  the  ring  rotating  at  the  same  speed. 

To  accommodate  this  difference  the  arm,  if  rigidly  fastened 
to  hub  and  rim,  will  be  extended  lengthwise  by  the  rim  and  the 
rim  will  be  drawn  in,  out  of  its  regular  circular  form,  by  the 
arm.  The  relation  between  the  amount  the  arm  is  drawn  out 
and  the  amount  the  rim  is  drawn  in  is  governed  by  the  propor- 
tions of  these  parts. 


FLY-WHEELS.  267 

The  result  is  that  the  rim  tends  to  bow  out  between  the  arms 
and  really  become  akin  to  a  uniformly  loaded  continuous  beam 
with  the  dangerous  sections  midway  between  the  arms  and  at 
the  points  of  junction  of  arms  and  rim.  The  fallacy  of  applying 
the  ring  theory  solely  to  the  fly-wheel  rim  becomes  evident  at 
once.  In  a  free  ring  the  form  of  cross-section  is  immaterial,  as 
the%section  is  subjected  only  to  tension.  In  the  rim  with  arms 
the  form  of  cross-section  becomes  a  vital  point,  as  the  rim  is 
subjected  to  flexure  as  well  as  tension,  and  the  strength  of  a 
member  to  resist  flexure  depends  directly  upon  the  modulus 
of  the  section. 

In  addition  to  the  foregoing  stresses,  which  are  induced 
under  all  conditions,  even  under  the  extreme  supposition  that 
the  wheel  is  rotating  at  a  perfectly  uniform  rate,  there  are  others 
when  the  rim  is  considered  as  performing  its  functions — i.e., 
in  a  balance-wheel,  absorbing  or  giving  out  energy  by  changes 
of  velocity  and,  in  a  band-wheel,  transmitting  the  power. 

This  may  be  seen  by  reference  to  Fig.  191.  A  shows  the 
relation  between  rim,  arm,  and  hub  when  the  wheel  is  at  rest 
or  rotating  uniformly  and  not  transmitting  ^ — i — ^  -^-"""T-^N^ 
any  power.  B  shows  the  relation  when 
work  is  being  done.  The  arm  becomes  an 
encastre  beam  and  corresponding  stresses 
are  induced  in  it.  Furthermore,  the  bend-  FIG.  191. 

ing  of  the  arm  tends  to  shorten  it  radially,  thus  drawing  in  the 
outer  end,  which  increases  the  flexure  in  the  rim.  In  addition 
to  the  foregoing  there  are  stresses  in  the  rim  due  to  the  weight 
of  the  wheel,  shrinkage,  etc.,  which  cannot  be  eliminated. 

180.  Stresses  in  Arms  of  Pulleys  or  Fly-wheels. — The  arms 
are  principally  stressed  by  the  bending  moment  due  to  variations 
of  velocity  of  the  wheel  or  to  the  power  transmitted. 

Let  T=the  greatest    turning   moment  transmitted  in  inch- 
pounds; 


268'  MACHINE  DESIGN. 

w=  number  of  arms; 

ft  =safe  unit  stress  in  outer  fiber  of  arm  in  pounds  per 
square  inch; 

—  =modulus  of  section  of  arm,  dimensions  in  inches. 
Then 

T=nft-  may  be  written  and  solved  for—. 
c  c 

Having  determined  upon  the  form  of  cross-section  the  dimensions 

can  be  determined  from  this  value  of—. 

c 

If  T  is  unknown  the  arms  can  be  made  as  strong  as  the  shaft 
by  equating  the  twisting  strength  of  the  shaft  to  the  bending 
s  trength  of  the  arms,  thus : 


£=  allowable  shearing  stress  in  outer  fiber  of  shaft,  pounds  per 

square  inch; 
r  =  radius  of  shaft  in  inches; 

n,  ft,  and  —  as  before. 

Consider  junction  of  arm  and  hub  next.     (See 

Fig.  192.) 
FIG.  192. 

The    tendency    for    the    arm    to  fail    through 

flexure  on  the  section    A- A    may  be  equated  to  the  tendency 
for  the  bolts  2  and  3  to  shear  off,  using  i  as  a  pivot. 

Let  A  =  combined  shearing  areas  of  2  and  3,  square  inches; 
/8=  allowable  shearing  stress  of  2  and  3  in  pounds  per 

square  inch; 

/=distance  between  centers  i  and  2,  and    i  and  3,  in 
inches; 


FLY-WHEELS.  269 

ft=  allowable  stress  in  outer  fiber  of  arm  in  pounds  per 
square  inch; 

—  =  modulus  of  arm  section,  dimensions  in  inches. 
c 


Then 


which  can  be  solved  for  A,  the  desired  area. 

If  the  arm  is  bolted  to  the  rim  a  similar  method  may  be 
employed  to  make  the  bolts  as  strong  as  the  arm. 

181.  Construction  of  Fly-wheels. — Since  weight  is  so  great 
a  factor  in  fly-wheels  it  has  been  common  practice  to  make 
them  of  that  material  which  combines  greatest  weight  with 
least  cost,  namely,  cast  iron.  That  this  is  not  always  safe 
practice  has  been  conclusively  demonstrated  by  many  serious 
accidents. 

Up  to  10  feet  in  diameter  the  wheels  are  generally  cast 
in  a  single  piece.  Occasionally  the  hub  is  divided  to  relieve 
the  stresses  due  to  cooling.  In  such  cases,  supposing  the  wheel 
to  have  six  arms,  the  hub  is  made  in  three  sections,  each  having 
a  pair  of  arms  running  to  the  rim.  Since  the  sections  are  inde- 
pendent, any  pair  of  arms  can  adjust  itself  to  the  conditions  of 
shrinkage  without  subjecting  the  other  arms  to  indeterminate 
stresses.  The  hub  sections  are  separated  from  each  other  by 
a  space  of  half  an  inch  or  less  and  this  is  filled  with  lead  or 
babbitt  metal.  Then  shrink-rings  or  bolts  are  used  to  hold 
the  sections  together.  Sometimes  the  hub  is  only  split  into  two 
parts. 

For  reasons  connected  chiefly  with  transportation,  wheels 
from  10  to  15  feet  in  diameter  are  cast  in  two  halves  which 
are  afterwards  joined  together  by  flanges  and  bolts  at  the  rim, 
and  shrink-rings  or  bolts  at  the  hub. 


270  MACHINE  DESIGN. 

In  still  larger  and  heavier  wheels  the  hub  is  generally  made 
entirely  separate  from  the  arms.  The  rim  is  made  in  as  many 
segments  as  there  are  arms.  Sometimes  the  arm  is  cast  with 
the  segment  and  sometimes  the  arms  and  segments  are  cast 
separately.  The  hub  is  commonly  made  in  the  form  of  a  pair 
of  disks  having  a  space  between  them  to  receive  the  arms  which 
are  fastened  to  them  by  means  of  accurately  fitted  through  bolts. 

Unless  the  wheel  is  to  be  a  forced  fit  on  its  shaft  it  is  best 
to  have  three  equally  spaced  keyways,  so  that  it  may  be  kept 
accurately  centered  with  the  shaft. 

In  these  large  wheels  the  joints  of  the  segments  of  the  rim 
are  usually  midway  between  the  arms  and  steel  straps  or  links 
"^^T^t  such  as  are  shown  in  Fig.  193  are 
A  ^  w-""i}^'  heated  and  dropped  into  recesses  pre- 
FlG-  I93>  viously  fitted  to  receive  them.  As  they 

cool,  their  contraction  draws  the  joint  together.  They  should 
not,  however,  be  subjected  to  a  very  great  initial  tension  of  this 
sort.  The  form  shown  at  A  is  most  commonly  used.  The  links 
are  made  of  high-grade  steel  and  their  area  is  such  that  their  ten- 
sile strength  equals  that  of  the  reduced  section  of  the  rim.  The 
areas  subjected  to  shear  and  compression  must  also  have  this 
strength. 

Taking  the  nature  of  the  stresses  into  consideration  it  is  clear 
that  the  rim  should  always  be  as  deep  radially  as  possible  to 
resist  the  flexure  action,  also  that  the  arms  should  be  near  together. 
Many  arms  are  much  better  than  a  few  and  a  disk  or  web  is  still 
better.* 

The  strongest  wheel  having  arms  will  be  one  whose  rim  is 
cast  in  a  single  piece,  while  the  arms  and  hubs  are  cast  as  a  second 
piece.  On  the  inside  of  the  rim  there  are  lugs  between  which 

*  Disk  wheels  have  the  further  advantage  of  offering  less  resistance  to  the 
air.  This  maybe  a  considerable  item.  See  Cassier's  Mag., Vol.  23,  pp.  577  and 
761. 


FLY-WHEELS.  271 

the  ends  of  the  arms  fit  so  that  there  is  a  space  of  about  one 
fourth  inch  all  around.  (See  Fig.  194.)  This  space  may  be 
filled  with  oakum  well  driven  in.  It  is  clear  that  the  rim  in  this 
case  acts  as  a  free  ring  and  is  subjected  solely  to  centrifugal 
tension.* 

Joints  in  the  rim  must  always  be  a  source  of  weakness  whether 
located  at  the  end  of  the  arms  or  midway  between  arms. 


FIG.  194.  FIG.  195. 

If  a  flanged  joint  midway  between  the  arms  is  used,  such  as 
is  shown  in  Fig.  195,  which  is  common  practice  for  split  band- 
wheels  of  medium  size,  the  flanges  should  be  deep  radially  and 
well  braced  by  ribs.  The  bolts  should  be  as  close  to  the  rim 
as  possible,  and  a  tension  rod  should  carry  the  extra  stress  (due 
to  the  weight  of  the  heavy  joint  and  its  velocity)  to  the  hub. 
Experiments  made  by  Prof.  C.  H.  Benjamin  f  show  that  the  use 
of  such  tie-rods  increases  the  strength  of  the  wheel  100  per  cent 
over  that  of  a  similar  wheel  without  tie-rods.  He  also  found 
that  jointed  rims  are  only  one  fourth  as  strong  as  solid  rims. 

Probably  as  strong  a  form  of  cast-iron  built-up  wheel  for 
heavy  duty  as  any  yet  designed  is  one  described  by  Mr.  John 
Fritz,!  having  a  hollow  rim  and  many  arms. 

But,  at  the  best,  cast  iron  is  an  uncertain  material  to  use 
for  such  tensile  and  flexure  stresses  as  are  induced  in  a  heavy- 
duty  fly-wheel,  and  it  is  wiser  to  make  such  wheels  of  structural 
steel.  A  built-up  wheel  having  a  disk  or  web  of  steel  plates  and 
a  rim  of  the  same  material,  all  joints  being  carefully  "broken  " 
and  strongly  riveted,  is  so  much  better  than  any  built-up  cast- 

*  See  Trans.  A.  S.  M.  E.,  Vol.  XX,  p.  944,  and  Vol.  XXI,  p.  322. 
t  Ibid.,  Vols.  XX  and  XXIII. 
j  Ibid.,  Vol.  XXI. 


272  MACHINE-DESIGN. 

iron  wheel  that  the  latter  are  passing  out  of  use.  The  steel 
wheels  can  have  at  least  twice  the  rim  velocity  of  the  cast  wheels 
with  greater  safety  and  may  therefore  be  much  lighter  for  the 
same  duty.  Their  lesser  wehht  makes  less  pressure  on  the 
bearings  and  consequently  le:s  friction  loss.* 

Plate  II  shows  forms  cf  rim  joints  for  split  rim  flywheels 
and  pulleys  which  are  probably  as  strong  as  any  that  can  be 
devised.  They  are  taken  from  the  American  Machinist,  Vol.  30. 
It  is  a  mistake,  however,  to  believe  th-t  any  of  these  joints  will 
will  give  as  strong  a  wheel  as  one  having  a  solid  rim. 

*  For  drawings  and  descriptions  of  wheels  made  of  forged  materials  the  reader 
is  referred  to  Vol.  XVII,  Trans.  A.  S.  M.  E.,  and  Power,  April  1894,  Nov.  i8o5> 
Jan.  1896,  and  Nov.  1897. 


PLATE  II. 


HAIGHT'S  JOINT  FOR  HEAVY  RIM. 


FLANGED  JOINT   OVER-ARM 
SEGWENTAL  RIM. 


DOUBLE  ARM  JOINT  FOR   WIDE  RIM. 


DOUBLE  ARM  JOINT  FOR  SHEAVE  WHEEL. 


CHAPTER  XVII. 

TOOTHED  WHEELS   OR   GEARS. 

182.  Fundamental  Theory  of  Gear  Transmission.  —  When 
toothed  wheels  are  used  to  communicate  motion,  the  motion 
elements  are  the  tooth  surfaces.  The  contact  of  these  surfaces 
with  each  other  is  line  contact.  Such  pairs  of  motion  elements 
are  called  higher  pairs,  to  distinguish  them  from  lower  .pairs, 
which  are  in  contact  throughout  their  entire  surface.  Fig.  196 
shows  the  simplest  toothed-wheel  mechanism.  There  are  three 
links,  a,  b,  and  c,  and  therefore  three  centres,  ab,  be,  and  ac.  These 
centres  must,  as  heretofore  explained,  lie  in  the  same  straight 
line,  ac  and  ab  are  the  centers  of  the  turning  pairs  connecting 
c  and  b  to  a.  It  is  required  to  locate  be  on  the  line  of  centers. 

When  the  gear  c  is  caused  to  rotate  uniformly  with  a  certain 
angular  velocity,  i.e.,  at  the  rate  of  m  revolutions  per  minute, 
it  is  required  to  cause  the  gear  b  to  rotate  uniformly  at  a  rate 
of  n  revolutions  per  minute.  The  angular  velocity  ratio  is  there- 

fore constant   and  =  —  .     The  centro  be  is  a  point  on  the  line  of 

centers  which  has  the  same  linear  velocity  whether  it  is  con- 
sidered as  a  point  in  &  or  c.  The  linear  velocity  of  this  point  be 
mb  =  zrcRin;  and  the  linear  velocity  of  the  same  point  in  c  =  27:R2m  ; 
in  which  RI  =  radius  of  be  in  b,  and  R2  =  radius  of  be  in  c.  But 
this  linear  velocity  must  be  the  same  in  both  cases,  and  hence  the 
above  expressions  may  be  equated  thus: 


275 


whence 


MACHINE  DESIGN. 
Rim 


Hence  be  is  located  by  dividing  the  line  of  centers  into  parts  which 

are  to  each  other  inversely  as  the  angular  velocities  of  the  gears. 

Thus,  let  ab  and  ac,  Fig.  197,  be  the  centers  of  a  pair  of  gears 

whose   angular  velocity  ratio  =—.      Draw  the   line   of    centers; 

divide  it  into  m  +  n  equal  parts;  m  of  these  from  ab  toward  the 
right,  or  n  from  ac  toward  the  left,  will  locate  be.  Draw  circles 
through  be,  with  ab  and  ac  as  centers.  These  circles  are  the 
centrodes  of  be  and  are  called  pilch  circles.  It  has  been  already 
explained  that  any  motion  may  be  reproduced  by  rolling  the 
centrodes  of  that  motion  upon  each  other  without  slipping. 


FIG.  196. 


FIG.  197. 


Therefore  the  motion  of  gears  is  the  same  as  that  which  would 
result  from  the  rolling  together  of  the  pitch  circles  (or  cylinders) 
without  slipping.  In  fact,  these  pitch  cylinders  themselves 
might  be,  and  sometimes  are,  used  for  transmitting  motion  of 
rotation.  Slipping,  however,  is  apt  to  occur,  and  hence  these 
"friction-gears"  cannot  be  used  if  no  variation  from  the  given 
velocity  ratio  is  allowable.  Hence  teeth  are  formed  on  the 
wheels  which  engage  with  each  other,  to  prevent  slipping. 

183.  Definitions. — If  the  pitch  circle  be  divided  into  as  many 
equal  parts  as  there  are  teeth  in  the  gear,  the  arc  included  between 


TOOTHED   WHEELS   OR   GEARS.  277 

two  of  these  divisions  is  the  circular  pitch  *  of  the  gear.  Circular 
pitch  may  also  be  defined  as  the  distance  on  the  pitch  circle 
occupied  by  a  tooth  and  a  space;  or,  otherwise,  it  is  the  distance 
on  the  pitch  circle  from  any  point  of  a  tooth  to  the  corresponding 
point  in  the  next  tooth.  A  fractional  tooth  is  impossible,  and 
therefore  the  circular  pitch  must  be  such  a  value  that  the  pitch 
circumference  is  divisible  by  it.  Let  P=  circular  pitch  in  inches; 
let  D=  pitch  diameter  in  inches;  N=  number  of  teeth;  then 

NP=xD;    N  =  ~;    D= ;    P=^T.     From   these   relations 

-t  71  J.\ 

any  one  of  the  three  values,  P,  D,  and  N,  may  be  found  if  the 
other  two  are  given. 

Diametral  pitch  is   the  number  of    teeth  per  inch  of   pitch 

N 
diameter.     Thus  if  />=diametral  pitch,  P=j^-      Multiplying  the 

nD  N  nD    N 

two  expressions,  P  =~TTF  and  p  =yr,  together  gives  Pp  =~TF  •  7T  =7r- 

Or,  the  product  of  diametral  and  circular  pitch  =x.  Circular 
pitch  is  usually  used  for  large  cast  gears,  and  for  mortise-gears 
(gears  with  wooden  teeth  inserted).  Diametral  pitch  is  usually 
used  for  small  cut  gears. 

In  Fig.  198,  b,  c,  and  k  are  pitch  points  of  the  teeth;  the  arc 
bk  is  the  circular  pitch;  ab  is  the  face  of  the  tooth;  bm  is  the  flank 
of  the  tooth ;  the  whole  curve  abm  is  the  profile  of 
the  tooth;  AD  is  the  total  depth  of  the  tooth;  AC 
is  the  working  depth;  AB  is  the  addendum;  a 
circle  through  A  is  the  addendum  circle.  Clear- 
ance is  the  excess  of  total  depth  over  working 
depth,  =CD.  Backlash  is  the  width  of  space  on  the  pitch  line 
minus  the  wridth  of  the  tooth  on  the  same  line.  In  cast  gears 
whose  tooth  surfaces  are  not  "tooled,"  backlash  needs  to  be 

*  Sometimes  called  circumferential  pitch. 


278  MACHINE  DESIGN. 

allowed,  because  of  unavoidable  imperfections  in  the  surfaces. 
In  cut  gears,  however,  it  may  be  reduced  almost  to  zero,  and  the 
tooth  and  space,  measured  on  the  pitch  circle,  may  be  considered 
equal. 

184.  Conditions  Governing  Forms  of  Teeth. — Teeth  of  almost 
any  form  may  be  used,  and  the  average  velocity  will  be  right. 
But  if  the  forms  are  not  correct  there  will  be  continual  variations 
of  velocity  ratio  between  a  minimum  and  maximum  value.     These 
variations  are  in  many  cases  unallowable,  and  in  all  cases  unde- 
sirable.    It  is  necessary  therefore  to  study  tooth  outlines  which 
shall  serve  for  the  transmission  of  a  constant  velocity  ratio. 

The  centro  of  relative  motion  of  the  two  gears  must  remai.t 
in  a  constant  position  in  order  that  the  velocity  ratio  shall  b-' 
constant.  The  essential  condition  jor  constant  velocity  ratio  is, 
therefore,  that  the  position  o]  ike  centro  o]  relative  motion  oj  the 
gears  shall  remain  unchanged.  If  A  and  B,  Fig.  199,  are  tooth 
surfaces  in  contact  at  a,  their  only  possible  relative  motion,  if 
they  remain  in  contact,  is  slipping  motion  along  the  tangent  CD. 
The  centro  of  this  motion  must  be  in  EF,  a  normal  to  the  tooth 
surfaces  at  the  point  of  contact.  If  these  be  supposed  to  be 
teeth  of  a  pair  of  gears,  b  and  c,  whose  required  velocity  ratio 
is  known,  ^nd  whose  centro,  be,  is  therefore  located,  then  in 
order  that  the  motion  communicated  from  one  gear  to  the  other 
through  the  point  of  contact,  a,  shall  be  the  required  motion,  it 
is  necessary  that  the  centro  of  the  relative  motion  of  the  teeth 
shall  coincide  with  be. 

185.  Illustration. — In  Fig.  200,  let  ac  and  ab  be  centers  of 
rotation  of  bodies  b  and  c,  and  the  required  velocity  ratio  is  such 
that  the  centro  of  b  and  c  falls  at  be.     Contact  between  b  and  c 
is  at  p.    The  only  possible  relative  motion  if  these  surfaces  re- 

'main  in  contact  is  slipping  along  CD;  hence  the  centro  of  this 
motion  must  be  on  EF,  the  normal  to  the  tooth  surfaces  at  the 
point  of  contact.  But  it  must  also  be  on  the  same  straight  line 


TOOTHED   WHEELS  OR   GEARS.  279 

with  ac  and  ab;  hence  it  is  at  be,  and  the  motion  transmitted  for 
the  instant,  at  the  point  p,  is  the  required  motion,  because  its 
centre  is  at  be.  But  the  curves  touching  at  p  might  be  of  such 
form  that  their  common  normal  at  p  would  intersect  the  line  of 
centers  at  some  other  point,  as  K,  which  would  then  become  the 
centre  of  the  motion  of  b  and  c  for  the  instant,  and  would  corre- 
spond to  the  transmission  of  a  different  motion.  The  essential 
condition  to  be  fulfilled  by  tooth  outlines,  in  order  that  a  con- 
stant velocity  ratio  may  be  maintained,  may  therefore  be  stated 
as  follows:  The  tooth  outlines  must  be  such  that  their  normal  at 
the  point  oj  contact  shall  always  pass  through  the  centra  corre- 
sponding to  the  required  'velocity  ratio. 

FIG.  199. 


FIG.  200. 

186.  Given  Tooth  Outline  to  Find  Form  of  Engaging  Tooth. 
— Having  given  any  curve  that  will  serve  for  a  tooth  outline 
in  one  gear,  the  corresponding  curve  may  be  found  in  the  other 
gear,  which  will  engage  with  the  given  curve  and  transmit  a 

constant   velocity   ratio.      Let  —   be   the    given    velocity   ratio. 

w+w=the  sum  of  the  radii  of  the  two  gears.  Draw  the  line 
of  centers  AB,  Fig.  201.  Let  P  be  the  "pitch  point,"  i.e.,  the 
point  of  contact  of  the  pitch  circles  or  the  centre  of  relative 
motion  of  the  two  gears.  To  the  right  from  P  lay  off  a  distance 
PB=m;  from  P  toward  the  left  lay  off  PA  =w,  A  and  B  will 
then  be  the  required  centers  of  the  wheels,  and  the  pitch  circles 


28o  MACHINE  DESIGN. 

may  be  drawn  through  P.  Let  abc  be  any  given  curve  on  the 
wheel  A.  It  is  required  to  find  the  curve  in  B  which  shall  engage 
with  abc  to  transmit  the  constant  velocity  ratio  required.  A 
normal  to  the  point  of  contact  must  pass  through  the  centre. 
If,  therefore,  any  point,  as  a,  be  taken  in  the  given  curve,  and  a 
normal  to  the  curve  at  that  point  be  drawn,  as  aa,  then  when  a 
is  the  point  of  contact,  a  will  coincide  with  P.  Also,  if  Cf  is  a 
normal  to  the  curve  at  c,  then  7-  will  coincide  with  P  when  c  is 
the  point  of  contact  between  the  gears;  and  since  b  is  in  the  pitch- 
line,  it  will  itself  coincide  with  P  when  it  is  the  point  of  contact. 

FIG.  201. 


FIG.  202. 

Since  the  two  pitch  circles  must  roll  upon  each  other  without 
slipping,  it  follows  that  the  arc  Pa'  =  arc  Pa,  arc  P&'=arc  Pb, 
and  arc  Pf  =  arc  Pf. 

Rotate  the  point  a,  about  A,  through  the  angle  0.  At  the 
same  time  a'  rotates  backward  about  B  through  the  angle  6r 
and  a  and  a'  coincide  at  P.  Pa"  represents  the  rotated  position 
of  the  normal  aa.  Rotate  Pa"  about  B  through  the  angle  0';. 
P  will  coincide  with  a'  and  a"  will  locate  the  point  a'  of  the  desired 
tooth  outline  of  gear  B.  The  point  V  of  the  desired  outline  is 
readily  located  by  merely  laying  off  arc  PV  =arc  Pb. 

(f  is  located  by  the  same  method  we  employed  to  determine 
a'.  This  will  give  three  points  in  the  required  curve,  and  through. 


TOOTHED   WHEELS  OR   GE4RS.      ,  281 

these  the  curve  may  be  drawn.    The  curve  could,  of  course,  be 
more  accurately  determined  by  using  more  points. 

Many  curves  could  be  drawn  that  would  not  serve  for  tooth 
outlines;  but,  given  any  curve  that  will  serve,  the  corresponding 
curve  may  be  found.  There  would  be,  therefore,  almost  an 
infinite  number  of  curves  that  would  fulfill  the  requirements  of 
correct  tooth  outlines.  But  in  practice  two  kinds  of  curves  are 
found  so  convenient  that  they  are  most  commonly,  though  not 
exclusively,  used.  They  are  cycloidal  and  involute  curves. 

187.  Cycloidal  Tooth  Outlines.— It  is  assumed  that  the  char- 
acter of  cycloidal  curves  and  method  of  drawing  them  is  under- 
stood. 

In  Fig.  202,  let  b  and  c  be  the  pitch  circles  of  a  pair  of  wheels, 
always  in  contact  at  be.  Also,  let  m  be  the  describing  circle  in 
contact  with  both  at  the  same  point.  M  is  the  describing  point. 
When  one  curve  rolls  upon  another,  the  centre  of  their  relative 
motion  is  always  their  point  of  contact.  For,  since  the  motion 
of  rolling  excludes  slipping,  the  two  bodies  must  be  stationary, 
relativp  to  each  other,  at  their  point  of  contact;  and  bodies  that 
move  relative  to  each  other  can  have  but  one  such  stationary 
point  in  common — their  centre.  When,  therefore,  m  rolls  in 
or  upon  b  or  c,  its  centre  relatively  to  either  is  their  point  of  con- 
tact. The  point  M,  therefore,  must  describe  curves  whose 
direction  at  any  point  is  at  right  angles  to  a  line  joining  that 
point  to  the  point  of  contact  of  m  with  the  pitch  circles.  Suppose 
the  two  circles  b  and  c  to  revolve  about  their  centers,  being  always 
in  contact  at  bc\  suppose  m  to  rotate  at  the  same  time  about  its 
center,  the  three  circles  being  always  in  contact  at  one  point  and 
having  no  slip.  The  point  M  will  then  describe  simultaneously 
a  curve,  bf)  on  the  plane  of  b,  and  a  curve,  c',  on  the  plane  of  c. 
Since  M  describes  the  curves  simultaneously,  it  will  always  be 
the  point  of  contact  between  them  in  any  position.  And  since 
the  point  M  moves  always  at  right  angles  to  a  line  which  joins  it 


282  MACHINE  DESIGN. 

to  be,  therefore  the  normal  to  the  tooth  surfaces  at  their  poirt 
of  contact  will  always  pass  through  be,  and  the  condition  for 
constant  velocity  ratio  transmission  is  fulfilled.  But  these  curves 
are  precisely  the  epicycloid  and  hypocycloid  that  would  be  drawn 
by  the  point  M  in  the  generating  circle,  by  rolling  on  the  out- 
side of  b  and  inside  of  c.  Obviously,  then,  the  epicycloids  and 
hypocycloids  generated  in  this  way,  used  as  tooth  profiles,  will 
transmit  a  constant  velocity  ratio. 

This  proof  is  independent  of  the  size  of  the  generating  circle, 
and  its  diameter  may  therefore  equal  the  radius  of  c.  Then  the 
hypocycloids  generated  by  rolling  within  c  would  be  straight 
lines  coinciding  with  the  radius  of  c.  In  this  case  the  flanks 
of  the  teeth  of  c  become  radial  lines,  and  therefore  the  teeth  are 
thinner  at  the  base  than  at  the  pitch  line;  for  this  reason  they 
are  weaker  than  if  a  smaller  generating  circle  had  been  used.  All 
tooth  curves  generated  with  the  same  generating  circle  will  work 
together,  the  pitch  being  the  same.  It  is  therefore  necessary 
to  use  the  same  generating  circle  for  a  set  of  gears  which  need 
to  interchange* 

The  describing  circle  may  be  made  still  larger.  In  the  first 
case  the  curves  described  have  their  convexity  in  the  same  direction, 
i.e.,  they  lie  on  the  same  side  of  a  common  tangent.  When  the 
diameter  of  the  describing  circle  is  made  equal  to  the  radius 
of  c,  one  curve  becomes  a  straight-line  tangent  to  the  other  curve. 
As  the  describing  circle  becomes  still  larger,  the  curves  have  their 
convexity  in  opposite  directions.  As  the  circle  approximates 
equality  with  c,  the  hypocycloid  in  c  grows  shorter,  and  finally 
when  the  describing  circle  equals  c,  it  becomes  a  point  which  is 
the  generating  point  in  c,  which  is  now  the  generating  circle.  If 
this  point  could  be  replaced  by  a  pin  having  no  sensible  diameter, 
it  would  engage  with  the  epicycloid  generated  by  it  in  the  other 
gear  to  transmit  a  constant  velocity  ratio.  But  a  pin  without 

*  See  §  ic^ 


TOOTHED   WHEELS  OR   GEARS.  283 

sensible  diameter  will  not  serve  as  a  wheel-tooth,  and  a  proper 
diameter  must  be  assumed,  and  a  new  curve  laid  off  to  engage 
with  it  in  the  other  gear.  In  Fig.  203,  AB  is  the  epicycloid  gener- 
ated by  a  point  in  the  circumference  of  the 
other  pitch  circle.  CD  is  the  new  curve 
drawn  tangent  to  a  series  of  positions  of  the 
pin  as  shown.  The  pin  will  engage  with 
this  curve,  CD,  and  transmit  the  constant 
velocity  ratio  as  required.  In  Fig.  202,  let  it 
be  supposed  that  when  the  three  circles  rotate 
constantly  tangent  to  each  other  at  the  pitch  point  be,  a  pencil  is 
fastened  at  the  point  M  in  the  circumference  of  the  describing 
circle.  If  this  pencil  be  supposed  to  mark  simultaneously  upon 
the  planes  of  b,  c,  and  that  of  the  paper,  it  will  describe  upon  b 
an  epicycloid,  on  c  a  hypocycloid,  and  on  the  plane  of  the 
paper  an  arc  of  the  describing  circle.  Since  M  is  always 
the  point  of  contact  of  the  cycloidal  curves  (because  it  gen- 
erates them  simultaneously),  therefore,  in  cycloidal  gear-teeth, 
the  locus  or  path  o)  the  point  of  contact  is  an  arc  0}  the  describ- 
ing circle.  The  ends  of  this  path  in  any  given  case  are  located 
by  the  points  at  which  the  addendum  circles  cut  the  describing 
circles. 

In  the  cases  already  considered,  where  an  epicycloid  in  one 
wheel  engages  with  a  hypocycloid  in  the  other,  the  contact  of 
the  teeth  with  each  other  is  all  on  one  side  of  the  line  of  centers. 
Thus,  in  Fig.  202,  if  the  motion  be  reversed,  the  curves  will  be 
in  contact  until  M  returns  to  be  along  the  arc  MD-bc;  but  after 
M  passes  be  contact  will  cease.  If  c  were  the  driving-wheel,  the 
point  of  contact  would  approach  the  line  of  centers;  if  b  were  the 
driving-wheel  the  point  of  contact  would  recede  from  the  line  of 
centers.  Experience  shows  that  the  latter  gives  smoother  running 
because  of  better  conditions  as  regards  friction  between  the  tooth 
surfaces.  It  would  be  desirable,  therefore,  that  the  wheel  with 
the  epicycloidal  curves  should  always  be  the  driver.  But  it 


284  MACHINE  DESIGN. 

should  be  possible  to  use  either  wheel  as  driver  to  meet  the  varying 
conditions  in  practice. 

Another  reason  why  contact  should  not  be  all  on  one  side  of 
the  line  of  centers  may  be  explained  as  follows. 

188.  Definitions:  Pitch-arc,  Arc  of  Action,  Line  of  Pressure. 
— The  angle  through  which  a  gear-wheel  turns  while  one  of  its 
teeth  is  in  contact  with  the  corresponding  tooth  in  the  other  gear 
is  called  the  angle  of  action.  It  is  found  by  drawing  radial  lines 
from  the  center  of  the  pitch  circle  to  the  two  ends  of  the  path  of 
action.  The  arc  of  the  pitch  circle  corresponding  to  the  angle 
of  action  is  called  the  arc  of  action. 

The  arc  of  action  must  be  greater  than  the  "pitch  arc"  (the 
arc  of  the  pitch  circle  that  includes  one  tooth  and  one  space),  or 
else  contact  will  cease  between  one  pair  of  teeth  before  it  begins 
between  the  next  pair.  Constrainment  would  therefore  not  be 
complete,  and  irregular  velocity  ratio  with  noisy  action  would 
result. 

In  Fig.  204,  let  AB  and  CD  be  the  pitch  circles  of  a  pair  of 
gears  and  E  the  describing  circle.  Let  an  arc  of  action  be  laid 
off  on  each  of  the  circles  from  P,  as  Pa,  PC,  and  Pe.  Through 
e,  about  the  center  O,  draw  an  addendum  circle*  i.e.,  the  circle 
which  limits  the  points  of  the  teeth.  Since  the  circle  E  is  the 
path  of  the  point  of  contact,  and  since  the  addendum  circle 
limits  the  points  of  the  teeth,  their  intersection,  e,  is  the  point  at 
which  contact  ceases,  rotation  being  as  indicated  by  the  arrow. 
If  the  pitch  arc  just  equals  the  assumed  arc  of  action,  contact 
will  be  just  beginning  at  P  when  it  ceases  at  e;  but  if  the  pitch 
arc  be  greater  than  the  arc  of  action,  contact  will  not  begin  at 
P  till  after  it  has  ceased  at  e,  and  there  will  be  an  interval  when 
AB  will  not  drive  CD.  The  greater  the  arc  of  action  the  greater 
the  distance  of  e  from  P  on  the  circumference  of  the  describing 
circle.  The  direction  of  pressure  between  the  teeth  is  always 
a  normal  to  the  tooth  surface,  and  this  always  passes  through 


TOOTHED   WHEELS  OR   GE/tRS. 


285 


the  pitch  point;  therefore  the  greater  the  arc  of  action — i.e., 
the  greater  the  distance  of  e  from  P — the  greater  the  obliquity 
of  the  line  of  pressure.  The  pressure  may  be  resolved  into  two 
components,  one  at  right  angles  to  the  line  of  centers  and  the 
other  parallel  to  it.  The  first  is  resisted  by  the  teeth  of  the  follower- 
wheel,  and  therefore  produces  pressure  with  resulting  friction. 
Hence  it  follows  that  the  greater  the  arc  of  action  the  greater 
will  be  the  average  obliquity  of  the  line  of  pressure,  and  there- 
fore the  greater  the  component  of  the  pressure  that  produces 


FIG.  204. 


FIG.  205. 


wasteful  friction.  If  it  can  be  arranged  so  that  the  arc  of  action 
shall  be  partly  on  each  side  of  the  line  of  centers,  the  arc  of  action 
may  be  made  greater  than  the  pitch  arc  (usually  equal  to  about 
1 1  times  the  pitch  arc);  then  the  obliquity  of  the  pressure-line 
may  be  kept  within  reasonable  limits,  contact  between  the  teeth 
will  be  insured  in  all  positions,  and  either  wheel  may  be  the 
driver.  This  is  accomplished  by  using  two  describing  circles  as 
in  Fig.  205.  Suppose  the  four  circles  A,  B,  a,  and  /?  to  roll  con- 
stantly tangent  at  P.  a  will  describe  an  epicycloid  on  the  plane 
of  B,  and  a  hypocycloid  on  the  plane  of  A.  These  curves  will 
engage  with  each  other  to  drive  correctly.  /?  will  describe  an 
epicycloid  on  A,  and  a  hypocycloid  on  B.  These  curves  will 
engage,  also,  to  drive  correctly.  If  the  epicycloid  and  hypocycloid 
in  each  gear  be  drawn  through  the  same  point  on  the  pitch  circle, 
a  double  curve  tooth  outline  will  be  located,  and  one  curve  will 


286  MACHINE  DESIGN. 

engage  on  one  side  of  the  line  of  centers  and  the  other  on  the 
other  side.  If  A  drives  as  indicated  by  the  arrow,  contact  will 
begin  at  D,  and  the  point  of  contact  will  follow  an  arc  of  a  to  P, 
and  then  an  arc  of  /?  to  C. 

189.  Involute  Tooth  Outlines. — If  a  string  is  wound  around  a 
cylinder  and  a  pencil-point  attached  to  its  end,  thi  s  point  will 
trace  an  involute  on  a  plane  normal  to  the  axis  of  the  cylinder 
as  the  string  is  unwound  from  the  cylinder.  Or,  if  the  point 
be  constrained  to  follow  a  tangent  to  the  cylinder,  and  the  string 
be  unwound  by  rotating  the  cylinder  about  its  axis,  the  point 
will  trace  an  involute  on  a  plane  that  rotates  with  the  cylinder. 

Illustration. — Let  a,  Fig.  206,  be  a  circular  piece  of  wood  free 
to  rotate  about  C;  /?  is  a  circular  piece  of  cardboard  made  fast 
to  a;  AB  is  a  straight-edge  held  on  the  circumference  of  a, 
having  a  pencil-point  at  B.  As  B  moves  along  the  straight-edge 
to  A,  a  and  /?  rotate  about  C,  and  B  traces  an  involute  DB  upon 
/?,  the  relative  motion  of  the  tracing  point  and  /?  being  just  the 
same  as  if  the  string  had  been  simply  unwound  from  a  fixed. 
If  the  tracing  point  is  caused  to  return  along  the  straight-edge 
it  will  trace  the  involute  BD  in  a  reverse  direction. 


FIG.  206.  FIG.  207. 

The  centro  of  the  tracing  point  is  always  the  point  of  tan- 
gency  of  the  string  with  the  cylinder;  therefore  the  string,  or 
straight-edge,  in  Fig.  208,  is  always  at  right  angles  to  the  direc- 
tion of  motion  of  the  tracing  point,  and  hence  is  always  a  normal 
to  the  involute  curve.  Let  a  and  3,  Fig.  207,  be  two  base  cylin- 
ders; let  AB  be  a  cord  wound  upon  a  and  /?  and  passing  through 


TOOTHED   WHEELS  OR   GEARS.  287 

the  centre  P,  which  corresponds  to  the  required  velocity  ratio. 
Let  a  and  /?  be  supposed  to  rotate  so  that  the  cord  is  wound  from 
/?  upon  a.  Then  any  point  in  the  cord  will  move  from  A  toward 
B,  and,  if  it  be  a  tracing-point,  will  trace  an  involute  of  a  on  the 
plane  of  a  (extended  beyond  the  base  cylinder) ,  and  will  also  trace 
an  involute  of  /?  upon  the  plane  of  /?.  These  two  involutes  will 
serve  for  tooth  profiles  for  the  transmission  of  the  required  con- 
stant velocity  ratio,  because  AB  is  the  constant  normal  to  both 
curves  at  their  point  of  contact,  and  it  passes  through  P,  th.3 
centre  corresponding  to  the  required  velocity  ratio.  Hence  the 
necessary  condition  is  fulfilled.  The  pitch  circles  will  have  OP 
and  O'P  as  their  respective  radii. 

Since  a  point  in  the  line  AB  describes  the  two  involute  curves 
simultaneously,  the  point  of  contact  of  the  curves  is  always  in 
the  line  AB.  And  hence  AB  is  the  path  of  the  point  of  contact. 
In  any  given  case  the  two  ends  of  the  path  lie  at  the  intersections 
of  the  addendum  circles  with  AB.  The  angle  of  action  and  the 
arc  0}  action  are  found  by  drawing  radial  lines  from  the  center 
of  the  pitch  circle  to  the  ends  of  the  path  of  contact. 

One  of  the  advantages  of  involute  curves  for  tooth  profiles 
is  that  a  change  in  distance  between  centers  of  the  gears  does 
not  interfere  with  the  transmission  of  a  constant  velocity  ratio. 
This  may  be  proved  as  follows:  In  Fig.  207,  from  similar  triangles 

O  7?       OP 

fy-j-  =ryp',   that  is,  the  ratio  of  the  radii  of  the  base  circles  (i.e., 

sections  of  the  base  cylinders)  is  equal  to  the  ratio  of  the  radii  of 
the  pitch  circles.  This  ratio  equals  the  inverse  ratio  of  angular 
velocities  of  the  gears.  Suppose  now  that  O  and  O'  be  moved 
nearer  together;  the  pitch  circles  will  be  smaller,  but  the  ratio 
of  their  radii  must  be  equal  to  the  unchanged  ratio  of  the 
radii  of  the  base  circles,  and  therefore  the  velocity  ratio  remains 
unchanged.  Also  the  involute  curves,  since  they  are  generated 
from  the  same  base  cylinders,  will  be  the  same  as  before,  and 


288  MACHINE  DESIGN. 

therefore,  with  the  same  tooth  outlines,  the  same  constant  velocity 
ratio  will  be  transmitted  as  before. 

190.  Racks. — A  rack  is  a  wheel  whose  pitch  radius  is  infinite; 
its  pitch  circle,  therefore,  becomes  a  straight  line,  and  is  tangent 
to  the  pitch  circle  of  the  wheel,  or  pinion,*  with  which  the  rack 
engages.  The  line  of  centers  is  a  normal  to  the  pitch  line  of  the 
rack,  through  the  center  of  the  pitch  circle  of  the  pinion.  The 
pitch  of  the  rack  is  determined  by  laying  off  the  circular  pitch 
of  the  engaging  wheel  on  the  pitch  line  of  the  rack.  The  curves 
of  the  cycloidal  rack-teeth,  like  those  of  wheels  of  finite  radius, 
may  be  generated  by  a  point  in  the  circumference  of  a  circle  which 
rolls  on  the  pitch  circle.  Since,  however,  the  pitch  circle  is  now 
a  straight  line,  the  tooth  curves  will  be  cycloids,  both  for  flanks 
and  faces.  In  Fig.  208,  AB  is  the  pitch  circle  of  the  pinion  and 
CD  is  the  pitch  line  of  the  rack;  a  and  b  are  describing  circles. 
Suppose,  as  before,  that  all  move  without  slipping  and  are  con- 
stantly tangent  at  P.  A  point  in  the  circumference  of  a  will  then 


FIG.  208. 

describe  simultaneously  a  cycloid  on  CD,  and  a  hypocycloid 
within  AB.  These  will  be  correct  tooth  outlines.  Also,  a  point 
in  the  circumference  of  b  will  describe  a  cycloid  on  CD  and  an 
epicycloid  on  AB.  These  will  be  correct  tooth  outlines.  To 
find  the  path  of  the  point  of  contact,  draw  the  addendum  circle 
EF  of  the  pinion,  and  the  addendum  line  GH  of  the  rack.  When 
the  pinion  turns  clockwise  and  drives  the  rack,  contact  will  begin 

*  Pinion  is  a  word  to  denote  a  gear  having  a  low  number  of  teeth,  or  the  smaller 
one  of  a  pair  of  engaging  gears. 


TOOTHED    WHEELS  OR   GEARS.  28g 

at  e  and  follow  arcs  of  the  describing  circles  through  P  to 
K.  It  is  obvious  that  a  rack  cannot  be  used  where  rotation 
is  continuous  in  one  direction,  but  only  where  motion  is 
reversed. 

Involute  curves  may  also  be  used  for  the  outlines  of  rack 
teeth.  Let  CD  and  C'D',  Fig.  209,  be  the  pitch  lines.  When  it 
is  required  to  generate  involute  curves  for  tooth  outlines,  for  a 
pair  of  gears  of  finite  radius,  a  line  is  drawn  through  the  pitch 
point  at  a  given  angle  to  the  line  of  centers  (usually  75°) ;  this 
line  is  the  path  of  the  point  which  generates  two  involutes  simul- 
taneously, and  therefore  the  path  of  the  point  of  contact  between 
the  tooth  curves.  It  is  also  the  common  tangent  to  the  two  base 
circles,  which  may  now  be  drawn  and  used  for  the  describing  of 
the  involutes.  To  apply  this  to  the  case  of  a  rack  and  pinion, 
draw  EF,  Fig.  209,  making  the  desired  angle  with  the  line  of 
centers,  OP.  The  base  circles  must  be  drawn  tangent  to  this 
line;  AB  will  therefore  be  the  base  circle  for  ih?.  pinion.  But 
the  base  circle  in  the  rack  has  an  infinite  radius,  and  a  circle  of 
infinite  radius  drawn  tangent  to  EF  would  be  a  straight  line 
coincident  with  EF.  Therefore  EF  is  the  base  line  of  the  rack. 
But  an  involute  to  a  base  circle  of  infinite  radius  is  a  straight 
line  normal  to  the  circumference — in  this  case  a  straight  line  per- 
pendicular to  EF.  Therefore  the  tooth  profiles  of  a  rack  in  the 
involute  system  will  always  be  straight  lines  perpendicular  to  the 
path  of  the  describing  point,  and  passing  through  the  pitch  points. 
If,  in  Fig.  209.  the  pinion  move  clockwise  and  drive  the  rack,  the 
contact  will  begin  at  E,  the  intersection  of  the  addendum  line 
of  the  rack  GH,  and  the  path  of  the  point  of  contact  EF,  and 
will  follow  the  line  EF  through  P  to  the  point  where  EF  cuts  the 
addendum  circle  LM  of  the  pinion. 

191.  Annular  Gears. — Both  centers  of  a  pair  of  gears  may  be 
on  the  same  side  of  the  pitch  point.  This  arrangement  corre- 
sponds to  what  is  known  as  an  annular  gear  and  pinion.  Thus, 


290 


MACHINE  DESIGN. 


in  Fig.  210,  AB  and  CD  are  the  pitch  circles,  and  their  centers  are 
both  above  the  pitch  point  P.  Teeth  may  be  constructed  to 
transmit  rotation  between  AB  and  CD.  AB  will  be  an  ordinary 


FIG.  211. 


spur  pinion,  but  it  is  obvious  that  CD  becomes  a  ring  of  metal 
with  teeth  on  the  inside,  i.e.,  it  is  an  annular  gear.  In  this  case 
a.  and  /?  may  be  describing  circles  for  cycloidal  teeth,  and  a  point 
in  the  circumference  of  a  will  describe  hypocycloids  simultaneously 
on  the  planes  of  AB  and  CD;  and  a  point  in  the  circumference 
of  /?  will  describe  epicycloids  simultaneously  on  the  planes  of  AB 
and  CD.  These  will  engage  to  transmit  a  constant  velocity 
ratio.  Obviously  the  space  inside  of  an  annular  gear  corresponds 
to  a  spur-gear  of  the  same  pitch  and  pitch  diameter,  with  tooth 
curves  drawn  with  the  same  describing  circle.  Let  EF  and  GH, 
Fig.  212,  be  the  addendum  circles.  If  the  pinion  move  clockwise 
driving  the  annular  gear,  the  path  of  the  point  of  contact  will  be 
from  e  along  the  circumference  of  a  to  P,  and  from  P  along  the 
circumference  of  /3  to  K. 

The  construction  of  involute  teeth  for  an  annular  gear  and 
pinion  involves  exactly  the  same  principles  as  in  the  case  of  a 
pair  of  spur- gears.  The  only  difference  of  detail  is  that  the 
describing  point  is  in  the  tangent  to  the  base  circles  produced 
instead  of  being  between  the  points  of  tangency.  Let  O  and  Of> 


TOOTHED   WHEELS  OR   GE4RS.  291 

Fig.  211,  be  the  centers,  and  AB  and  //  the  pitch  circles  of  an 
annular  gear  and  pinion.  Through  P,  the  point  of  tangency  of 
the  pitch  circles,  draw  the  path  of  the  point  of  contact  at  the  given 
angle  with  the  line  of  centers.  With  O  and  O'  as  centers  draw 
tangent  circles  to  this  line.  These  will  be  the  involute  base 
circles.  Let  the  tangent  be  replaced  by  a  cord,  made  fast,  say, 
at  K',  winding  on  the  circumference  of  the  base  circle  CK',  to  D, 
and  then  around  the  base  circle  FE  in  the  direction  of  the  arrow, 
and  passing  over  the  pulley  G,  which  holds  it  in  line  with  PB. 
If  rotation  be  supposed  to  occur  with  the  two  pitch  circles  always 
tangent  at  P  without  slipping,  any  point  in  the  cord  beyond  P 
toward  G  will  describe  an  involute  on  the  plane  77,  and  another 
on  the  plane  of  AB.  These  will  be  the  correct  involute  tooth 
profiles  required.  Draw  NQ  and  LM,  the  addendum  circles. 
Then  if  the  pinion  move  clockwise,  driving  the  annular  gear, 
the  point  of  contact  starts  from  e  and  moves  along  the  line  GH 
through  P  to  K. 

When  a  pair  of  spur-gears  mesh  with  each  other,  the  direction 
of  rotation  is  reversed.  But  an  annular  gear  and  pinion  meshing 
together  rotate  in  the  same  direction. 

192.  Interchangeable  Sets  of  Gears. — In  practice  it  is  desirable 
to  have  "  interchangeable  sets  "  of  gears;  i.e.,  sets  in  which  any 
gear  will  "mesh  "  correctly  with  any  other,  from  the  smallest 
pinion  to  the  rack,  and  in  which,  except  for  limiting  conditions 
of  size,  any  spur-gear  will  mesh  with  any  annular  gear.  Inter- 
changeable sets  may  be  made  in  either  the  cycloidal  or  involute 
system.  A  necessary  condition  in  any  set  is  that  the  pitch  shall 
be  constant,  because  the  thickness  of  tooth  on  the  pitch  line  must 
always  equal  the  width  of  the  space  (less  backlash).  If  this 
condition  is  unfulfilled  they  cannot  engage,  whatever  the  form 
of  the  tooth  outlines. 

The  second  condition  for  an  interchangeable  set  in  the 
cycloidal  system  is  that  the  size  of  the  describing  circle  shall  be 


292  MA CHINE  DESIGN. 

constant.  If  the  diameter  of  the  describing  circle  equal  the  radius 
of  the  smallest  pinion's  pitch  circle,  the  flanks  of  this  pinion's 
teeth  will  be  radial  lines,  and  the  tooth  will  therefore  be  thinner 
at  the  base  than  at  the  pitch  line.  As  the  gears  increase  in  size 
with  this  constant  size  of  describing  circle,  the  teeth  grow  thicker 
at  the  base;  hence  the  weakest  teeth  are  those  of  the  smallest 
pinion. 

It  is  found  unadvisable  to  make  a  pinion  with  less  than  twelve 
teeth.  If  the  radius  of  a  fifteen-tooth  pinion  be  selected  for  the 
diameter  of  the  describing  circle,  the  flanks  which  bound  a  space 
in  a  twelve-tooth  pinion  will  be  very  nearly  parallel,  and  may 
therefore  be  cut  with  a  milling-cutter.  This  would  not  be  possi- 
ble if  the  describing  circle  were  made  larger,  causing  the  space 
to  become  wider  at  the  bottom  than  at  the  pitch  circle.  There- 
fore the  maximum  describing  circle  for  milled  gears  is  one  whose 
diameter  equals  the  pitch  radius  of  a  fifteen -tooth  pinion  and 
it  is  the  one  usually  selected.  Each  change  in  the  number  of 
teeth  with  constant  pitch  causes  a  change  in  the  size  of  the  pitch 
circle.  Hence  the  form  of  the  tooth  outline,  generated  by  a 
describing  circle  of  constant  diameter,  also  changes.  For  any 
pitch,  therefore,  a  separate  cutter  would  be  required  corresponding 
to  every  number  of  teeth,  to  insure  absolute  accuracy.  Practi- 
cally, however,  this  is  not  necessary.  The  change  in  the  form 
of  tooth  outline  is  much  greater  in  a  small  gear,  for  any  increase 
in  the  number  of  teeth,  than  in  a  large  one.  It  is  found  that 
twenty-four  cutters  will  cut  all  possible  gears  of  the  same  pitch 
with  sufficient  practical  accuracy.  The  range  of  these  cutters 
is  indicated  in  the  following  table,  taken  from  Brown  and  Sharpe's 
"Treatise  on  Gearing." 

These  same  principles  of  interchangeable  sets  of  gears  with 
cycloidal  tooth  outlines  apply  not  only  to  small  milled  gears  as 
above,  but  also  to  large  cast  gears  with  tooled  or  untooled  tooth 
surfaces. 


TOOTHED   WHEELS    OR   GEARS 
TABLE  XXVI. 


293 


A  cuts  12  teeth                       Cutter  M  cuts    27  to 

29  teeth 

B     "     13     ' 

< 

N     ' 

3°  " 

33     " 

C     "    14     ' 

1 

O    ' 

34  " 

37     ' 

D    "     15     ' 

' 

P 

38" 

42     ' 

E     "     16     ' 

* 

Q    ' 

43  " 

49     ' 

F     "     17     ' 

-  « 

R     ' 

5°  " 

59     ' 

G    "     18     < 

' 

S      ' 

60  " 

74     ' 

H    "     19     ' 

' 

T     ' 

75  " 

99     ' 

I        "      20       '                                               ' 

U     ' 

100   " 

149     ' 

J     "    21  to  22  teeth                 4 

V     ' 

150  " 

249     " 

K    "    23  "  24     "                    ' 

W    ' 

250  " 

rack 

L    "    24  "  26     "                            X     '      rack 

193.  Interchangeable  Involute  Gears. — In  the  involute  system 
the  second  condition  of  interchangeability  is  that  the  angle  between 
the  common  tangent  to  the  base  circles  and  the  line  of  centers  shall 
be  constant.  This  may  be  shown  as  follows:  Draw  the  line  of 
centers,  AB,  Fig.  212.  Through  P,  the  assumed  pitch  point, 
draw  CD,  and  let  it  be  the  constant  common  tangent  to  all  base 


FIG.  212. 

circles  from  which  involute  tooth  curves  are  to  be  drawn.  Draw 
any  pair  of  pitch  circles  tangent  at  P,  with  their  centers  in  the 
line  AB.  About  these  centers  draw  circles  tangent  to  CD;  these 
are  base  circles,  and  CD  may  represent  a  cord  that  wind's  from 
one  upon  the  other.  A  point  in  this  cord  will  generate  simul- 
taneously involutes  that  will  engage  for  the  transmission  of  a  con- 
stant velocity  ratio.  But  this  is  true  of  any  pair  of  circles  that 
have  their  centers  in  AB,  and  are  tangent  to  CD.  Therefore, 


2Q4 


MACHINE  DESIGN. 


if  the  pitch  is  constant,  any  pair  of  gears  that  have  the  base  circles 
tangent  to  the  line  CD  will  mesh  together  properly.  As  in  the 
cycloidal  gears,  the  involute  tooth  curves  vary  with  a  variation 
in  the  number  of  teeth,  and,  for  absolute  theoretical  accuracy, 
there  would  be  required  for  each  pitch  as  many  cutters  as  there 
are  gears  with  different  numbers  of  teeth.  The  variation  is 
least  at  the  pitch  line,  and  increases  with  the  distance  from  it. 
The  involute  teeth  are  usually  used  for  the  finer  pitches  and  the 
cycloidal  teeth  for  the  coarser  pitches;  and  since  the  amount  that 
the  tooth  surface  extends  beyond  the  pitch  line  increases  with  the 
pitch,  it  follows  that  the  variation  in  form  of  tooth  curves  is 
greater  in  the  coarse  pitch  cycloidal  gears  than  in  the  fine  pitch 
involute  gears.  For  this  reason,  with  involute  gears,  it  is  only 
necessary  to  use  eight  cutters  for  each  pitch.  The  range  is  shown 
in  the  following  table,  which  is  also  taken  from  Brown  and  Sharpe's 
"Treatise  on  Gearing": 

TABLE  XXVII. 
No.  i  will  cut  wheels  from  135  teeth  to  racks 


'       2      " 

'        "          ' 

55 

'      "134  inclusive 

3 

35 

54 

'     4    " 

26 

34 

'    5    " 

'        "          ' 

21 

' 

25 

'    6    " 

"          ' 

17 

'      ' 

20        ' 

'    7    " 

<          :•            < 

14 

'      ' 

16 

8 

<           ti 

12 

'•     < 

T3         ' 

194.  Laying  Out  Gear-teeth.  —  Exact  and  Approximate 
Methods. — There  is  ordinarily  no  reason  why  an  exact  method 
for  kying  out  cycloidal  or  involute  curves  for  tooth  outlines  should 
not  be  used,  either  for  large  gears  or  gear  patterns,  or  in  making 
drawings."  It  is  required  to  lay  out  a  cycloidal  gear.  The  pitch, 
and  diameters  of  pitch  circle  and  describing  circle  are  given. 

Draw  the  pitch  circle  on  the  drawing-paper,  using  a  fine  line. 
On  a  flat  piece  of  tracing-cloth  or  thin,  transparent  celluloid  draw 
a  circle  the  size  of  the  generating  circle.  Use  a  fine,  clear  line. 


TOOTHED   WHEELS   OR.   GEARS  295 

Place  it  over  the  drawn  pitch  circle  so  that  it  is  tangent  to  the 
latter  at  P  as  shown  in  Fig.  213.  AB  is  an  arc  of  the  pitch  circle. 
Insert  a  needle  point  at  P,  and  using  it  as  a  pivot  swing  the 
tracing-cloth  in  the  direction  of  the -arrow  a  very  short  distance, 
so  that  the  generating  circle  cuts  the  pitch  circle  at  a  new  point 
Q,  as  shown  exaggeratedly  in  Fig.  214.  Q  should  be  taken  very  close 


FIG.  214.  FIG.  215.  FIG.  216. 

to  P.  Insert  a  needle  point  at  Q,  remove  the  one  at  P,  and  swing 
the  cloth,  about  Q  as  a  pivot,  in  the  direction  of  the  arrow  until 
the  two  circles  are  tangent  at  Q.  (See  Fig.  215.)  The  point  P 
of  the  tracing-cloth  now  lies  off  the  pitch  circle  a  short  distance. 
With  a  needle  point  prick  its  present  position  through  to  the 
drawing-paper.  Now  with  Q  as  a  pivot  rotate  the  tracing-cloth 
until  the  two  circles  intersect  at  a  point  R  slightly  beyond  Q. 
Insert  needle  at  R  and  remove  the  one  at  Q.  Swing  tracing- 
cloth  about  R  until  R  becomes  the  point  of  tangency  of  the  two 
circles  and  then  prick  the  new  position  of  P  through  to  the  drawing- 
paper.  Taking  points  very  near  together  and  repeating  the 
operation  gives  a  close  approximation  to  true  rolling  of  the  gen- 
erating circle  on  the  pitch  circle  and  therefore  the  path  of  the 
point  P  as  marked  on  the  drawing-paper  by  pricked  points  is  an 
epicycloid  and  may  be  used  for  the  face  of  the  tooth. 

Next  place  the  tracing-cloth  on  the  inside  of  the  pitch  circle, 
as  shown  in  Fig.  216,  with  the  generating  circle  tangent  to  the 
pitch  circle  at  the  original  point  P.  Using  the  method  just 
described  to  prevent  slipping,  roll  the  generating  circle,  in  the 
direction  of  the  arrow,  on  the  pitch  circle  and  the  path  traced 
by  P  as  marked  by  pricked  points  on  the  drawing-paper  will  be 
a  hypocycloid  for  the  flank  of  the  tooth. 


296 


MACHINE  DESIGN. 


The  compound  curve  aPb,  Fig.  217,  has  now  been  traced, 
which  forms  the  basis  of  the  completed  tooth  outline. 

AB  is  an  arc  of  the  pitch  circle  whose  center  is  at  O.  With 
O  as  center,  swing  in  the  addendum  circle  'CD  and  the  full 
depth  circle  EF,  according  to  the  proportions  given  in  §  195. 
With  a  radius  equal  to  TV  of  the  circular  pitch  draw  the  fillet 
cd  tangent  to  EF  and  aPb.  The  completed  tooth  profile  is  the 
curve  cdPe. 

Cut  a  wooden  template  to  fit  the  tooth  curve,  and  make  it 
fast  to  a  wooden  arm  free  to  rotate  about  O,  making  the  edge  of 
the  template  coincide  with  cdPe.  It  may  now  be  swung  succes- 
sively to  the  other  pitch  points,  and  the  tooth  outline  may  be  drawn 
by  the  template  edge.  This  gives  one  side  of  all  of  the  teeth. 
The  arm  may  now  be  turned  over  and  the  other  sides  of  the 
teeth  may  be  drawn  similarly. 


\  \ 


J  J 


D/ 

FIG.  217. 

It  is  required  to  lay  out  exact  involute  teeth.  The  pitch, 
pitch-circle  diameter,  and  angle  of  the  common  tangent  are 
given.— Draw  the  pitch  circle  AB,  Fig.  218,  and  the  line  of 
centers  OOr.  Through  the  pitch  point  P  draw  CD,  the  common 
tangent  to  the  base  circles,  making  the  angle  /?  with  the  line  of 
centers.  Draw  the  base  circle  EF  about  O  tangent  to  CD. 

On  a  piece  of  flat  tracing-cloth  draw  a  fine,  clear,  straight  line 
and  lay  the  tracing-cloth  over  the  drawing  so  that  this  line  coin- 
cides with  CD.  Take  a  needle  point  and  insert  it  at  the  point 


TOOTHED   WHEELS   OR   GEARS.  297 

of  tangency  Q.  With  another  needle,  mark  the  point  P  on  the 
tracing-cloth.  Now,  employing  a  pair  of  needle  points  to  pre- 
vent slipping,  roll  the  traced  line  on  the  base  circle  EF,  prick- 
ing the  path,  aPb,  of  the  point  P  of  the  tracing-cloth  through  to 
the  drawing-paper.  This  path  is  the  involute  of  the  base  circle 
and  is  the  basis  of  the  involute  tooth  outline.  To  complete  the 
latter  proceed  as  follows:  Draw  the  addendum  circle  GH  and 
the  full-depth  circle  JK.  In  general  JK  will  lie  inside  of  the 
base  circle  EF  and  it  will  be  necessary  to  extend  the  tooth  outline 
inward  beyond  a.  About  O  swing  a  circle  whose  diameter  equals 
one  half  the  circular  pitch  and  draw  ac  tangent  to  it.  With  a 
radius  equal  to  TV  of  the  circular  pitch  swing  in  the  fillet  de  tan- 
gent to  JK  and  ac.  deaPf  is  the  completed  outline.  If  the 
gear  has  20  teeth  or  less  ac  should  be  made  a  radial  line.  If 
EF  lies  inside  of  JK  we  draw  the  fillet  tangent  to  JK  and  aP* 

195.  Gear  Proportions. — The  following  formulas  and  table 
are  given  to  assist  in  the  practical  proportioning  of  gears : 
Let  D  =  pitch  diameter; 
D\  =  outside  diameter; 

D2=  diameter  of  a  circle  through  the  bottom  of  spaces; 
P  =  circular  pitch  =  space  on  the  pitch  circle  occupied  by 

a  tooth  and  a  space; 

p  =diametral  pitch  =  number  of  teeth  per  inch  of  pitch- 
circle  diameter; 
N  =numbcr  of  teeth; 
/  =  thickness  of  tooth  on  pitch  line; 
a  =  addendum; 

*  Approximate  tooth  outlines  may  be  drawn  by  the  use  of  instruments,  such 
as  the  Willis  odontograph,  which  locates  the  centers  of  approximate  circular  arcs; 
the  templet  odontograph,  invented  by  Prof.  S.  W.  Robinson;  or  by  some  geomet- 
rical or  tabular  method  for  the  location  of  the  centers  of  approximate  circular 
arcs.  For  descriptions,  see  "Elements  of  Mechanism,"  Willis;  "Kinematics," 
McCord;  "Teeth  of  Gears,"  George  B.  Grant;  "Treatise  on  Gearing,"  pub- 
lished by  Brown  and  Sharpe. 


298 


MACHINE  DESIGN. 


c  =  clearance; 

d  =  working  depth  of  spaces; 

di  =full  depth  of  spaces. 


Then 


N  +  2 


•T;   p=^; 

N 


D2=D-2(a+c); 
PN 


p      _ 

»  =  -fr;    P  =• — ;    t  =  — =— :,  no  backlash; 
P  '  p '          2      2p' 


t          P          T. 

c= — = —  =— — ; 

IO      20       P2O 


a 


—  inches. 


The  following  dimensions  are  given  as  a  guide;  they  may  be 
varied  as  conditions  of  design  require:  Width  of  face  =  about  ^P; 
thickness  of  rim=i.25/;  thickness  of  arms  =  i.25/,  no  taper. 
The  rim  may  be  reinforced  by  a  rib,  as  shown  in  Fig.  219.  Diam- 
eter of  hub  =  2Xdiameter  of  shaft.  Length  of  hub  =  width  of 
face  +  £";  width  of  arm  at  junction  with  hub=|  circumference 
of  the  hub  for  six  arms.  Make  arms  taper  about  \"  per  foot  on 
each  side. 

TABLE  XXVIII. 


Diametral 
Pitch. 

Circular  Pitch. 

Thickness  of 
Tooth  on  the 
Pitch  Line. 

Diametral 
Pitch. 

Circular  Pitch. 

Thickness  of 
Tooth  on  the 
Pitch  Line. 

P 

P 

< 

P 

P 

t 

1 

6.283 

3-I4I 

si 

.897 

•449 

1 

4.189 

2.094 

4 

•785 

•393 

I 

3-I4I 

I-57I 

5 

.628 

•314 

if 

•513 

1.256 

6 

•523 

.262 

15 

.094 

1.047 

7 

•  449 

.224 

jf 

•795 

•897 

8 

•393 

.196 

2 

•571 

•785 

9 

•349 

.174 

ii 

•396 
.256 

.142 

.698 
.628 

•571 

IO 
12 
14 

•  314 
.262 
.224 

•157 

.112 

3 

.047 

•523 

TOOTHED    WHEELS   OR   GE4RS.  299 

196.  Strength  of  Gear- teeth. — The  maximum  work  trans- 
mitted by  a  shaft  per  unit  time  may  usually  be  accurately  estimated ; 
and,  if  the  rate  of  rotation  is  known,  the  torsional  moment  may 
be  found.  Let  O,  Fig.  220,  represent  the  axis  of  a  shaft  perpen- 
dicular to  the  paper.  Let  A  =  maximum  work  to  be  transmitted 
per  minute;  let  N  =  revolutions  per  minute;  let  Fr  =  torsional 
moment.  Then  F  is  the  force  factor  of  the  work  transmitted, 

and  27trN  is  the  space  factor  of  the  work  transmitted.  ,  Hence 

^ 
zFnrN  =  A ,     and      Fr  =  torsional  moment  =  — ^. 

If  the  work  is  to  be  transmitted  to  another  shaft  by  means  of 
a  spur-gear  whose  radius  is  ri,  then  for  equilibrium  Firi=Fr, 

and  FI= — .     FI  is  the  force  at  the  pitch  surface  of  the  gear 

whcs?  radius  is  r\,  i.e.,  it  is  the  force  to  be  sustained  by  the  gear- 
teeth.  Hence,  in  general,  the  force  sustained  by  the  teeth  0}  a  gear 
equals  the  torsional  moment  divided  by  the  pitch  radius  of  the  gear. 

When  the  maximum  force  to  be  sustained  is  known  the  teeth 
may  be  given  proper  proportions.  The  dimensions  upon  which 
the  tooth  depends  for  strength  are :  Thickness  of  tooth  =/,  width 
of  face  of  gear  =  b,  and  depth  of  space  between  teeth  =  /.  These 
all  become  known  when  the  pitch  is  known,  because  t  is  fixed  for 
any  pitch,  and  /  and  b  have  values  dictated  by  good  practice. 
The  value  of  b  may  be  varied  through  quite  a  range  to  meet  the 
requirements  of  any  special  case. 

In  the  design  the  tooth  will  be  treated  as  a  cantilever  with 
a  load  applied  at  its  end.  It  is  assumed  that  one  tooth  sustains 
the  entire  load;  i.e.,  that  there  is  contact  only  between  one  pair 
of  teeth.  This  would  be  nearly  true  for  gears  with  low  numbers 
of  teeth ;  but  in  high-numbered  gears  the  force  would  be  distributed 
over  several  pairs,  and  hence  they  would  have  an  excess  of  strength. 
It  is  also  assumed  that  the  load  is  uniformly  distributed  across 
the  face  of  the  tooth.  This  is  a  safe  assumption  if  the  width  of 


500  MACHINE  DESIGN. 

face,  b,  does  not  exceed  three  times  the  circular  pitch,  i.e.,  $P,  and 
if  the  gears  are  well  aligned  and  rigidly  supported.  All  teeth  of 
the  same  pitch  have  not  the  same  form,  as  was  explained  in  the 
discussion  of  interchangeable  gears,  and  therefore  they  vary  in 
strength.  The  fewer  teeth  the  thinner  they  will  be  at  their  base 
and  consequently  the  weaker  they  will  be  when  acting  as  canti- 
levers. 


FIG.  219.  FIG.  220.  FIG   221. 

Mr.  Wilfred  Lewis  *  has  drawn  a  number  of  figures  on  a 
large  scale  to  represent  very  accurately  the  teeth  cut  by  complete 
sets  of  cutters  of  the  15°  involute,  the  20°  involute,  and  the  cycloidal 
systems.  In  the  latter  he  used  a  rolling  circle  having  a  diameter 
equal  to  the  radius  of  the  i2-tooth  pinion.  The  proportions 
of  the  teeth  used  in  his  investigation  are  slightly  different  from 
those  given  above  which  correspond  to  the  Brown  and  Sharpe 
system,  but  no  serious  errors  will  result  from  applying  the  formulas 
derived  by  him.  His  reasoning  was  as  follows  (see  Fig.  221): 

The  greatest  stress  in  the  tooth  occurs  when  the  load  is  applied 
at  the  end  of  the  tooth  as  indicated  by  the  arrow  at  a,  its  direction 
of  action  being  normal  to  the  profile  at  a.  The  component  of 
this  force,  which  is  effective  to  produce  rotation  of  the  gear,  equals 
F  and  is  called  the  working  force. 

This  load  is  applied  at  b  and  induces  a  transverse  stress  in 
the  tooth.  To  determine  where  the  tooth  is  weakest  advantage 
is  taken  of  the  fact  that  any  parabola  in  the  axis  be  and  tangent  to 

*  Proc.  Phila.  Eng.  Club,  1893,  and  Amer.  Mach.,  May  4  and  June  22,  1893. 


TOOTHED   WHEELS   OR  GEARS. 


bF  incloses  a  beam  of  uniform  strength.  Of  all  the  parabolas 
that  may  thus  be  drawn  one  only  will  be  tangent  to  the  tooth  form 
(as  shown  by  the  dotted  line  in  the  figure)  and  the  weakest  sec- 
tion of  the  tooth  will  be  that  through  the  points  of  tangency  c 
and  d.  Having  determined  the  weakest  section  in  each  case,  Mr. 
Lewis  developed  the  following  general  formulae  from  the  data  so 
obtained : 

For  1 5°  involute   and  the   cycloidal  system,  using   a   rolling 
circle  whose  diameter  equals  the  radius  of  the'i2-tooth  pinion, 


0.124 


0.684  \ 
'    N  )' 


For  the  20°  involute  system 


F=  working  force  in  pounds; 

/  =  safe  allowable  unit  stress  in  pounds  per  square  inch; 
P=  circular  pitch  in  inches; 
b  =  width  of  face  of  gear  in  inches ; 
N  =  number  of  teeth  in  the  gear. 

Experimental  data  fixing  the  value  of  /  for  different  materials 
and  velocities  are  lacking.  The  following  table  is  recommended 
by  Mr.  Lewis : 

TABLE  XXIX. 


Velocity  of   pitch  line 

in  feet  per  minute   .  . 

100  or 

200 

300 

600 

900 

1200 

1800 

24OO 

less 

/  for  cast  iron  

8000 

6000 

4800 

4000 

3000 

2400 

2000 

I70J 

/  for  steel  

20000 

15000 

12000 

10000 

7500 

6000 

5000 

4300 

*  For  the  cycloidal  system,  using 
radius  of  the  i5-tooth  pinion, 


rolling  circle  whose  diameter   equals  the 


(Trans  A.  S.  M   E.,  Vol.  18.  p.  776.) 


302  MACHINE  DESIGN. 

In  these  formulae,  for  a  given  gear  the  whole  right  side  of 
the  equation  becomes  known  and  the  allowable  value  of  F  is 
readily  determined.  It  is  more  difficult  to  apply  the  formulas 
where  the  force  to  be  transmitted  is  given.  In  such  a  case  the 
value  of  P  is  determined  by  trial. 

197.  Problem. — Design  a  pair  of  15°  involute  gears  to  transmit 
6  H.P.  The  distance  between  centers  is  10  inches  and  the 
velocity  ratio  of  the  shafts  is  to  be  f.  The  pinion  shaft  makes 
150  revolutions  per  minute. 

The  distance  between  centers  being  10  inches  and  the  velocity 
ration  §,  the  radii  will  be  to  each  other  as  3:2  and  their  sum, 
=  10  inches,  hence  the  radius  of  the  pinion  will  be  4  inches,  while 
that  of  the  gear  will  be  6  inches.  If  both  gears  are  of  the  same 
material  the  teeth  of  the  smaller  will  be  the  weaker.  Com- 
putations will  therefore  be  made  for  the  pinion  because  the  gear 
will  be  stronger  and,  consequently,  safe.  The  pitch  diameter 
of  the  pinion  =8  inches,  its  velocity  =  150X71  XT\=  3 14. 2  feet  per 
minute. 

6  H.P.  =6X33,000  =  198,000  ft. -Ibs.  per  minute; 

198000 

.F— —     —  =6^2  Ibs. 
314.2 

Assuming  cast-iron  gears  at  314.2  feet  per  minute,  /  may 
be  taken  =  4800  Ibs.  The  maximum  value  of  b  =  $P,  but  for 
first  trial  let  6  =  1  inch.  For  the  15°  involute  system,  then, 

/  o.684\ 

632=48ooXPXi(o.i24-— ^    ). 

"Riif  AT"        K  25-I3 

N=-p-=— p^. 

Substituting  this,  transposing,  and  clearing  of  fractions, 
P2-  4. 56^+4.84=0, 


TOOTHED   WHEELS  OR   GE4RS.  303 

from  which  P  —  2.28  =  ±c.6. 

Choosing  the  smaller  value, 

P  =  1.68  inches. 

A  pitch  of  1.68  inches  with  a  width  of  face  of  i  inch  is  bad  pro- 
portion. 

For  our  second  trial  let  6=2  inches, 

„/             o.684P\ 
632=4800X2X^(0.124- —  ). 

Solving  this  gives 

P=o.62  inch. 

This  corresponds  practically  to  a  diametral  pitch  of  5,  giving 
5X8=40  teeth  for  the  pinion.  The  engaging  gear  will  have 
5X12  =60  teeth. 

When  the  solution  of  the  quadratic  equation  leads  to  an 
imaginary  quantity  it  will  be  necessary  to  increase  b.  If  the 
greatest  allowable  value  of  b  still  leaves  the  imaginary,  then  the 
value  /  must  be  increased  either  by  using  a  stronger  material 
or  by  cutting  down  the  factor  of  safety. 

Another  way  of  solving  the  problem  would  have  been  to  assume 
b  in  terms  of  P.  thus  b  =  KP  (K  being  less  than  3),  thus  giving 


In  this   substitute  trial  values  for  P  until  one  is  found  which 
satisfies  the  equation. 

198.  Non-circular  Wheels. — Only  circular  centrodes  or  pitch 
curves  correspond  to  a  constant  velocity  ratio;  and  by  making 
the  pitch  curves  of  proper  form,  almost  any  variation  in  the  velocity 


304  MACHINE  DESIGN. 

ratio  may  be  produced.  Thus  a  gear  whose  pitch  curve  is  an 
ellipse,  rotating  about  one  of  its  foci,  may  engage  with  another 
elliptical  gear,  and  if  the  driver  has  a  constant  angular  velocity 
the  follower  will  have  a  continually  varying  angular  velocity. 
If  the  follower  is  rigidly  attached  to  the  crank  of  a  slider-crank 
chain,  the  slider  will  have  a  quick  return  motion.  This  is  some- 
times used  for  shapers  and  slotting-machines.  When  more  than 
one  fluctuation  of  velocity  per  revolution  is  required,  it  may  be 
obtained  by  means  of  "lobed  gears";  i.e.,  gears  in  which  the 
curvature  of  the  pitch  curve  is  several  times  reversed.  If  a 
describing  circle  be  rolled  on  these  non -circular  pitch  curves,  the 
tooth  outlines  will  vary7  in  different  parts;  hence  in  order  to  cut 
such  gears,  many  cutters  would  be  required  for  each  gear.  Prac- 
tically this  would  be  too  expensive;  and  when  such  gears  are 
used  the  pattern  is  accurately  made,  and  the  cast  gears  are  used 
without  "tooling"  the  tooth  surfaces. 

199.  Bevel-gears. — All  transverse  sections  of  spur-gears  arc 
the  same,  and  their  axes  intersect  at  infinity.  Spur-gears  serve 
to  transmit  motion  between  parallel  shafts.  It  is  necessary  also 
to  transmit  motion  between  shafts  whose  axes  intersect.  In  this 
case  the  pitch  cylinders  become  pitch  cones;  the  teeth  are  formed 
6  upon  these  conical  surfaces,  the  result - 
ing  gears  being  called  bevel-gears. 
To  illustrate,  let  a  and  b,  Fig.  222,  be 
the  axes  between  which  the  motion  is 
to  be  transmitted  with  a  given  velocity 
FlG  ratio.  This  ratio  is  equal  to  the  ratio 

of  the  length  of  the  line  A  to  that  of  B. 

Draw  a  line  CD  parallel  to  a,  at  a  distance  from  it  equal  to 
the  length  of  the  line  A.  Also  draw  the  line  CE  parallel  to  b, 
at  a  distance  from  it  equal  to  the  length  of  the  line  B.  Join 
the  point  of  intersection  of  these  lines  to  the  point  O,  the 
intersection  of  the  given  axes.  This  locates  the  line  OF,  which 


TOOTHED   WHEELS  OR   GEARS.  303 

is  the  line  of  contact  of  two  pitch  cones  which  wrill  roll  together 

K/f (~*        A 

to  transmit  the  required   velocity  ratio.     For  j^,  =  — ,  and  if  it 

be  supposed  that  there  are  frusta  of  cones  so  thin  that  they  may 
be  considered  cylinders,  their  radii  being  equal  to  MC  and  NC, 
it  follows  that  they  would  roll  together,  if  slipping  be  prevented, 
to  transmit  the  required  velocity  ratio.  But  all  pairs  of  radii 

MC 
of  these  pitch  cones  have  the  same  ratio,  =j^t  and  therefore 

any  pair  of  frusta  of  the  pitch  cones  may  be  used  to  roll  together 
for  the  transmission  of  the  required  velocity  ratio.  To  insure 
this  result,  slipping  must  be  prevented,  and  hence  teeth  are 
formed  upon  the  selected  frusta  of  the  pitch  cones.  The  theo- 
retical determination  of  these  may  be  explained  as  follows: 

200.  ist.  Cycloidal  Teeth.— If  a  cone,  A  (Fig.  223),  be  rolled 
upon  another  cone,  B,  their  apexes  coinciding,  an  element  be  of 
the  cone  A  will  generate  a  conical  surface,  and  a  spherical  sec- 


tion of  this  surface,  adb,  is  called  a  spherical  epicycloid.  Also 
if  a  cone,  A  (Fig.  224),  roll  on  the  inside  of  another  cone,  C, 
their  apexes  coinciding,  an  element  be  of  A  will  generate  a  conical 
surface,  a  spherical  section  of  which,  bda,  is  called  a  spherical 
hypocycloid.  If  now  the  three  cones,  B,  C,  and  A,  with  apexes 
coinciding,  roll  together,  always  tangent  to  each  other  on  one 
line,  as  the  cylinders  were  in  the  case  of  spur-gears,  there  will 
be  two  conical  surfaces  generated  by  an  element  of  A :  one  upon 
the  cone  B  and  another  upon  the  cone  C.  These  may  be  used 
for  tooth  surfaces  to  transmit  the  required  constant  velocity 


306  MACHINE  DESIGN. 

fatio.  Because,  since  the  line  of  contact  of  the  cones  is  the  axo  * 
of  the  relative  motion  of  the  cones,  it  follows  that  a  plane  normal 
to  the  motion  of  the  describing  element  of  the  generating  cone 
at  any  time  will  pass  through  this  axo.  And  also,  since  the 
describing  element  is  always  the  line  of  contact  between  the  gen- 
erated tooth  surfaces,  the  normal  plane  to  the  line  of  contact  of 
the  tooth  surfaces  always  passes  through  the  axo,  and  the  condi- 
tion of  rotation  with  a  constant  velocity  ratio  is  fulfilled. 

201.  2d.  Involute  Teeth. — If  two  pitch  cones  are  in  contact 
along  an  element,  a  plane  may  be  passed  through  this  element 
making  an  angle  (say  75°)  with  the  plane  of  the  axes  of  the  cones/ 
Tangent  to  this  plane  there  may  be  two  cones  whose  axes  coin- 
cide with  the  axes  of  the  pitch  cones.     If  a  plane  is  supposed  to 
wind  off  from  one  base  cone  upon  the  other,  the  line  of  tangency 
of  the  plane  with  one  cone  will  leave  the  cone  and  advance  in  the 
plane  toward  the  other  cone,  and  will  generate  simultaneously 
upon  the  pitch  cones'  conical  surfaces,  and  spherical  sections  of 
these  surfaces  will  be  spherical  involutes.     These  surfaces  may 
be  used  for  tooth  surfaces,  and  will  transmit  the  required  con- 
stant velocity  ratio,  because  the  tangent  plane  is  the  constant 
normal  to  the  tooth  surfaces  at  their  line  of  contact,  and  this 
plane  passes  through  the  axo  of  the  pitch  cones. 

202.  Determination  of  Bevel-gear  Teeth. — To  determine  the 
tooth  surfaces  with  perfect  accuracy,  it  would  be  necessary  to 
draw  the  required  curves  on  a  spherical  surface,  and  then  to 
join  all  points  of  these  curves  to  the  point  of  intersection  of  the 
axes  of  the  pitch  cones.     Practically  this  would  be  impossible, 
and  an  approximate  method  is  used. 

If  the  frusta  of  pitch  cones  be  given,  B  and  C,  Fig.  225,  then 
points  in  the  base  circles  of  the  cones,  as  L,  M,  and  K,  will  move 
always  in  the  surface  of  a  sphere  whose  projection  is  the  circle 


*  An  axo  is  an  instantaneous  axis,  of  which  a  centro  is  a  projection. 


TOOTHED   WHEELS  OR   GE4RS. 


307 


LA  KM.  Properly,  the  tooth  curves  should  be  laid  out  on  the 
surface  of  this  sphere  and  joined  to  the  center  of  the  sphere  to 
generate  the  tooth  surfaces.  Draw  cones  LGM  and  MHK  tan- 
gent to  the  sphere  on  circles  represented  in  projection  by  lines 
LM  and  MK.  They  are  called  the  "back  cones."  If  now 
tooth  curves  are  drawn  on  these  cones,  with  the  base  circle 
of  the  cones  as  pitch  circles,  they  will  approximate  the  tooth 
curves  that  should  be  drawn  on  the  spherical  surface.  But  a 
cone  may  be  cut  along  one  of  its  elements  and  rolled  out,  or  de- 
veloped, upon  a  plane.  Let  MDH  be  a  part  of  the  cone  MHK, 
developed,  and  let  MNG  be  a  part  of  the  cone  MGL,  developed. 


A 


FIG.  225. 


FIG.  226. 


The  circular  arcs  MD  and  MN  may  be  used  just  as  pitch  circles 
are  in  the  case  of  spur-gears,  and  the  teeth  may  be  laid  out  in 
exactly  the  same  way,  the  curves  being  either  cycloidal  or  in- 
volute, as  required.  Then  the  developed  cones  may  be  wrapped 
back  and  the  curves  drawn  may  serve  as  directrices  for  the 
tooth  surfaces,  all  of  whose  elements  converge  to  the  center  of 
the  sphere  of  motion. 

203.  Cutting  Bevel-gear  Teeth. — The  teeth  of  spur-gears  may 
be  cut  by  means  of  milling-cutters,  because  all  transverse  sections 
are  alike,  but  with  bevel-gears  the  conditions  are  different.  The 
tooth  surfaces  are  conical  surfaces,  and  therefore  the  curvature 
varies  constantly  from  one  end  of  the  tooth  to  the  other.  Also  the 


308  MACHINE  DESIGN. 

thickness  of  the  tooth  and  the  width  of  space  vary  constantly 
from  one  end  to  the  other.  But  the  curvature  and  thickness 
of  a  milling-cutter  cannot  vary,  and  therefore  a  milling-cutter 
cannot  cut  an  accurate  bevel-gear.  Small  bevel-gears  are, 
however,  cut  with  milling-cutters  with  sufficient  accuracy  for 
practical  purposes.  The  cutter  is  made  as  thick  as  the  narrowest 
part  of  the  space  between  the  teeth,  and  its  curvature  is  made 
that  of  the  middle  of  the  tooth.  Two  cuts  are  made  for  each 
space.  Let  Fig.  226  represent  a  section  of  the  cutter.  For  the 
first  cut  it  is  set  relatively  to  the  gear  blank,  so  that  the  pitch 
point  a  of  the  cutter  travels  toward  the  apex  of  the  pitch  cone, 
and  for  the  second  cut  so  that  the  pitch  point  b  travels  toward 
the  apex  of  the  pitch  cone.  This  method  gives  an  approximation 
to  the  required  form.  Gears  cut  in  this  manner  usually  need 
to  be  filed  slightly  before  they  work  satisfactorily.  Bevel-gears 
with  absolutely  correct  tooth  surfaces  may  be  made  by  planing. 
Suppose  a  planer  in  which  the  tool  point  travels  always  in  some 
line  through  the  apex  of  the  pitch  cone.  Then  suppose  that  as  it 
is  slowly  fed  down  the  tooth  surface,  it  is  guided  along  the  required 
tooth  curve  by  means  of  a  templet.  From  what  has  preceded 
it  will  be  clear  that  the  tooth  so  formed  will  be  correct.  Planers 
embodying  these  principles  have  been  designed  and  constructed 
by  Mr.  Corliss  of  Providence,  and  Mr.  Gleason  of  Rochester, 
with  the  most  satisfactory  results. 

204.  Design  of  Bevel-gears. — Given  energy  to  be  transmitted, 
rate  of  rotation  of  one  shaft,  velocity  ratio,  and  angle  between 
axes;  to  design  a  pair  of  bevel-gears.  Locate  the  intersection 
of  axes,  O,  Fig.  227.  Draw  the  axes  OA  and  OB,  making  the 
required  angle  with  each  other.  Locate  OC,  the  line  of  tangency 
of  the  pitch  cones,  by  the  method  given  on  p.  304.  Any  pair  of 
frusta  of  the  pitch  cones  may  be  selected  upon  which  to  form 
the  teeth.  Special  conditions  of  the  problem  usually  dictate  this 
selection  approximately.  Suppose  that  the  inner  limit  of  the 


TOOTHED   WHEELS  OR   GEARS. 


3°9 


teeth  may  be  conveniently  at  D.  Then  make  DP,  the  width 
of  face,  =DO+2.  Or,  if  P  is  located  by  some  limiting  condi- 
tion, lay  off  PD=PO+$.  In  either  case  the  limits  of  the  teeth 
are  denned  tentatively.  Now  from  the  energy  and  the  number 
of  revolutions  of  one  shaft  (either  shaft  may  be  used)  the  moment 
of  torsion  may  be  found.  The  mean  force  at  the  pitch  surface 
=  this  torsional  moment -f- the  mean  radius  of  the  gear;  i.e., 
the  radius  of  the  point  M,  Fig.  227,  midway  between  P  and  D. 
The  pitch  corresponding  to  this  force  may  be  found. 

In  order  to  compute  it  consider  the  teeth  of  the  pinion  (i.e., 
the  smaller  gear),  as  they  will  be  the  weaker.  Having  found 
the  force,  F,  which  is  to  be  transmitted  we  determine  the  pitch 
required  to  carry  F  by  a  spur-gear,  whose  pitch  radius  =M N 
and  whose  width  of  face,  b,  equals  PD.  (The  radius  MN  is 
used  to  govern  the  shape  of  the  tooth  and  not  MR,  because  the 
teeth  are  laid  off  on  the  developed  back  cones  and  not  on  the 
pitch  circles,  as  has  been  explained.  The  circle  whose  radius 
is  MN  is  sometimes  called  the  formative  circle  in  order  to  dis- 
tinguish it  from  the  pitch  circle.) 

The  pitch  of  such  a  spur-gear  would  be  the  mean  pitch  of  the 
bevel-gears.  But  the  pitch  of  bevel-gears  is  measured  at  the 
large  end,  and  diametral  pitch  varies  inversely  as  the  distance 
from  O.  In  this  case  the  distances  of  M  and  P  from  O  are  to 
each  other  as  5  is  to  6.  Hence  the  value  of  diametral  pitch  found 
Xf  =the  diametral  pitch  of  the  bevel-gear.  If  this  value  does 
not  correspond  with  any  of  the  usual  values  of  diametral  pitch, 
the  next  smaller  value  may  be  used.  This  would  result  in  a 
slightly  increased  factor  of  safety.  If  the  diametral  pitch  thus 
found,  multiplied  by  the  diameter  2PQ,  does  not  give  an  integer 
for  the  number  of  teeth,  the  point  P  may  be  moved  outward 
along  the  line  OC,  until  the  number  of  teeth  becomes  an  integer. 
This  also  would  result  in  slight  increase  of  the  factor  of  safety.  The 
point  P  is  thus  finally  located,  the  corrected  width  of  face  =  PO  -^3, 


MACHINE  DESIGN. 


and  the  pitch  is  known.  The  drawing  of  the  gears  may  be  com- 
pleted as  follows:  Draw  AB  perpendicular  to  PO.  With  A  and 
B  as  centers,  draw  the  arcs  PE  and  PF.  Use  these  as  pitch  arcs, 
and  draw  the  outlines  of  two  or  three  teeth  upon  them,  with 


FIG.  227. 

cycloidal  or  involute  curves  as  required.  These  will  serve  to  show 
the  form  of  tooth  outlines.  From  P  each  way  along  the  line  AB 
lay  off  the  addendum  and  the  clearance.  From  the  four  points 
thus  located  draw  lines  toward  O  terminating  in  the  line  DG. 
The  tops  of  teeth  and  the  bottoms  of  spaces  are  thus  denned. 
Lay  off  upon  AB  below  the  bottoms  of  the  spaces  a  space  HJ 
about  equal  to  |-.  the  thickness  of  the  tooth  on  the  pitch  circle. 
This  gives  a  ring  of  metal  to  support  the  teeth.  From  /  draw 
JK  toward  O.  The  web  L  should  have  about  the  same  thickness 
as  the  ring  has  at  K.  Join  this  web  to  a  properly  proportioned 
hub  as  shown.  The  plan  and  elevation  of  each  gear  may  now 
be  drawn  by  the  ordinary  methods  of  projection.  Use  large 
fillets. 


TOOTHED   WHEELS  OR   GEARS.  311 

205.  Skew  Bevel-gears. — Spur-gears  serve  to  communicate 
motion  between  parallel  axes,  and  bevel-gears  between  axes  that 
intersect.  But  it  is  sometimes  necessary  to  communicate  motion 
between  axes  that  are  neither  parallel  nor  intersecting.  If  the 
parallel  axes  are  turned  out  of  parallelism,  or  if  intersecting 
axes  are  moved  into  different  planes,  so  that  they  no  longer  inter- 
sect, the  pitch  surfaces  become  hyperbolic  paraboloids  in  contact 
with  each  other  along  a  straight  line,  which  is  the  generatrix  of 
the  pitch  surfaces.  These  hyperbolic  paraboloids  rolled  upon 
each  other,  circumferential  slipping  being  prevented,  will  trans- 
mit motion  with  a  constant  velocity  ratio.  There  is,  however, 
necessarily  a  slipping  of  the  elements  of  the  surfaces  upon  each 
other  parallel  to  themselves.  Teeth  may  be  formed  on  these  pitch 
surfaces,  and  they  may  be  used  for  the  transmission  of  motion 
between  shafts  that  are  not  parallel  nor  in  the  same  plane.  Such 
gears  are  called  "Skew  Bevel-gears."  The  difficulties  of  con- 
struction and  the  additional  friction  due  to  slipping  along  the 
elements  make  them  undesirable  in  practice,  and  there  is  seldom 
a  place  where  they  cannot  be  replaced  by  some  other  form  of 
connection. 

A  very  complete  discussion  of  the  subject  of  Skew  Bevel-gears 
may  be  found  in  Professor  McCord's  "Kinematics." 

206  Spiral  Gearing. — If  line  contact  is  not  essential  there  is 
much  wider  range  of  choice  of  gears  to  connect  shafts  which  are 
neither  parallel  nor  intersecting.  A  and  B,  Fig.  228,  are  axes  of 
rotation  in  different  planes,  both  planes  being 
parallel  to  the  paper.  Let  EF  and  GH  be  cylin- 
ders on  these  axes,  tangent  to  each  other  at  the 
point  S.  Any  line  may  now  be  drawn  through 
.  5  either  between  A  and  B  or  coinciding  with  either 
of  them.  This  line,  say  DSt  may  be  taken  as  the  common  tan- 
gent to  helical  or  screw  lines  drawn  on  the  cylinders  EF  and  GH, 
or  helical  surfaces  may  be  formed  on  both  cylinders,  DS  being 


312  MACHINE  DESIGN. 

their  common  tangent  at  S.  Spiral  gears  are  thus  produced. 
Each  one  is  a  portion  of  a  many-threaded  screw.  The  contact 
in  these  gears  is  point  contact;  in  practice  the  point  of  contact 
becomes  a  very  limited  area.* 

207.  Worm-gearing. — When  the  angle  between  the  shafts  is 
made  equal  to  90°,  and  one  gear  has  only  one,  two,  three,  or  four 
threads,  it  becomes  a  special  case  of  spiral  gearing  known  as 
Worm-gearing.  In  this  special  case  the  gear  with  a  few  threads 
is  called  the  worm,  while  the. other  gear,  which  is  still  a  many- 
threaded  screw,  is  called  the  worm-wheel.  If  a  section  of  a  worm 
and  worm-wheel  be  made  on  a  plane  passing  through  the  axis 
of  the  worm,  and  normal  to  the  axis  of  the  worm-wheel,  the  form 
of  the  teeth  will  be  the  same  as  that  of  a  rack  and  pinion;  in 
fact  the  worm,  if  moved  parallel  to  its  axis,  would  transmit  rotary 
motion  to  the  worm-wheel.  From  the  consideration  of  racks 
and  pinions  it  follows  that  if  the  involute  system  is  used,  the  sides 
of  the  worm-teeth  will  be  straight  lines.  This  simplifies  the  cutting 
of  the  worm,  because  a  tool  may  be  used  capable  of  being  sharpened 
without  special  methods.  If  the  worm-wheel  were  only  a  thin 
plate  the  teeth  would  be  formed  like  those  of  a  spur-gear  of  the 
same  pitch  and  diameter.  But  since  the  worm-wheel  must  have 
greater  thickness,  and  since  all  other  sections  parallel  to  that 
through  the  axis  of  the  worm,  as  CD  and  AB,  Fig.  229,  show  a 
different  form  and  location  of  tooth,  it  is  necessary  to  make 
the  teeth  of  the  worm-wheel  different  from  those  of  a  spur-gear, 
if  there  is  to  be  contact  between  the  worm  and  worm-wheel  any- 
where except  in  the  plane  EF,  Fig.  229.  This  would  seem  to 
involve  great  difficulty,  but  it  is  accomplished  in  practice  as 
follows :  A  duplicate  of  the  worm  is  made  of  tool  steel,  and  "  flutes  " 
are  cut  in  it  parallel  to  the  axis,  thus  making  it  into  a  cutter, 


*  See  further  "  Worm  and  Spiral  Gearing,"  by  F.  A.  Halsey.     Van  Nostrand 
Science  Series. 


TOOTHED   WHEELS  OR   GEARS. 


313 


which  is  tempered.  It  is  then  mounted  in  a  frame  in  the  same 
relation  to  the  worm-wheel  that  the  worm  is  to  have  when  they 
are  finished  and  in  position  for  working.  The  distance  between 
centers,  however,  is  somewhat  greater,  and  is  capable  of  being 
gradually  reduced.  Both  are  then  rotated  with  the  required 
velocity  ratio  by  means  of  gearing  properly  arranged,  and  the 
cutter  or  "hob  "  is  fed  against  the  worm-wheel  till  the  distance 
between  centers  becomes  the  required  value.  The  teeth  of  the 


FIG.  230. 


worm-wheel  are  "roughed  out"  before  they  are  "hobbed."  By 
the  above  method  the  worm  is  made  to  cut  its  own  worm- 
wheel.* 

Fig.  230  represents  the  half  section  of  a  worm.  If  it  is  a 
single  worm  the  thread  A,  in  going  once  around,  comes  to  B\ 
twice  around  to  C,  and  so  on.  If  it  is  a  double  worm  the  thread 
A,  in  going  once  around,  comes  to  C,  while  there  is  an  inter- 
mediate thread,  B.  It  follows  that  if  the  single  worm  turns 
through  one  revolution  it  will  push  one  tooth  of  the  worm-wheel 
with  which  it  engages  past  the  line  of  centers;  while  the  double 
worm  will  push  two  teeth  of  the  worm-wheel  past  the  line  of 
centers.  The  single  worm,  therefore,  must  make  as  many  revolu- 
tions as  there  are  teeth  in  the  worm-wheel,  in  order  to  cause 
one  revolution  of  the  worm-wheel;  while  for  the  same  result 
the  double  worm  only  needs  to  make  half  as  many  revolutions 


*  This  subject  is  fully  treated  in  Unwin's  "Elements  of   Machine  Design,' 
and  in  Brown  and  Sharpe's  "Treatise  on  Gearing." 


314  MACHINE  DESIGN. 

The  ratio  of  the  angular  velocity  of  a  single  worm  to  that  of 

the  worm-wheel  with  which  it  engages  is=— ,  in  which  n  equals 

the  number  of  teeth  in  the  worm-wheel.     For  the  double  worm 

n 
this  ratio  is  — . 

Worm-gearing  is  particularly  well  adapted  for  use  where  it 
is  necessary  to  get  a  high  velocity  ratio  in  limited  space. 

The  pitch  of  a  worm  is  measured  parallel  to  the  axis  of  rota- 
tion. The  pitch  of  a  single  worm  is  P,  Fig.  230.  It  is  equal 
to  the  circular  pitch  of  the  worm-wheel.  The  pitch  of  a  double 
worm  is  PI  =  2P  =  2Xcircular  pitch  of  the  worm-wheel. 

208.  Design  of  Worm-gears. — All  spiral  gears  are  forms  of 
screw  transmission  and  the  formulae  for  efficiency,  etc.,  developed 
under  c,  sec.  98,  in  the  chapter  on  Screws,  apply  to  them  directly. 

Three  points  are  to  be  carefully  considered  in  the  design  of 
worms  and  wheels: 

1.  Speed  o]  rubbing.     This  is  the  velocity  in  feet  per  minute 
of  a  point  on  the  pitch  line  of  the  worm.     The  best  efficiencies 
are  obtained  when  this  is  about  200  feet  per  minute.     When 
it  exceeds  300  feet  there  is  increasing  danger  of  cutting,  and 
the   pressure   on   the   teeth   must   be   correspondingly   reduced. 
At  high  speeds  (say  1000  feet)  only  very  light  pressure  can  be 
sustained  without  abrasion. 

2.  Pressure  on  teeth.     This  depends  on  the  speed  and  on  the 
angle  of  helix. 

3.  Angle  0}  helix.    From  the  formula  for  screw  efficiency  we 
have  seen  that  this  should  be  made  as  great  as  convenient  provided 
it  does  not  exceed  45°.     Practical  conditions  make  it  impossible 
to  use  the  highest  values,  but  20°  gives  very  excellent  results.    It 
should  never  be  less  than  15°  for  fair  efficiency. 

Oil-bath  lubrication  should  be  used  wherever  possible;  failing 
this,  a  heavy  mixture  of  graphite  and  oil  has  been  found  satis- 


TOOTHED   WHEELS  OR   GEARS. 


factory.  The  following  table  based  on  Professor  Stribeck's 
experiments  *  applies  to  a  20°  angle  of  helix  and  oil-bath  lubri- 
cation, using  a  hardened-steel  worm  and  phosphor-bronze 
wheel. 


TABLE  XXX. 


Rubbing  velocity  in  feet  per  minute  
Allowable  pressure  for  maximum  efficiency 
in  pounds  

200 

350° 

300 
2700 

400 
1850 

500 
1250 

600 

1000 

This  may  be  taken  as  a  guide.  When  the  angle  is  greater 
than  20°  the  values  of  the  pressure  may  be  slightly  increased. 
When  the  angle  is  less  than  20°  they  should  be  rapidly  diminished ; 
thus  for  10°  use  only  one  half  the  value  given. 

209.  Problem. — Two  shafts  about  10  inches  apart  and  at  a 
right  angle  with  one  another  are  to  have  a  velocity  ratio  of  20  to  i. 
The  worm-shaft  makes  300  revolutions  per  minute. 

Since  the  velocity  ratio  is  20  to  i,  the  wheel  will  have  to  have 
20,  40,  or  60  teeth,  depending  upon  whether  the  worm  is  single-, 
double-,  or  triple-threaded. 

If  the  shafts  are  10  inches  apart  the  greatest  allowable  pitch 
radius  of  the  wheel  will  not  be  far  from  8  inches ;  50  inches  may 
be  taken  as  a  trial  pitch  circumference  of  the  wheel. 

With  a  single-thread  worm  this  will  give  a  circular  pitch  of 
f$  =  2j  inches.  With  a  double  thread  the  circular  pitch  would 
be  |$  =i\  inches. 

In  any  case  the  rise  of  the  pitch  helix  of  the  worm  will  be 
2|  inches  for  one  revolution. 

This  value  must  always  be  such  that  the  thread  may  be  cut  in 
an  ordinary  lathe. 


Zeitschrift  d.  Vereins  deutscher  Ingenieure. 


316  MACHINE  DESIGN. 

If  it  is  required  that  the  helix  angle,   a,  be  20°,  then  the 
pitch  circumference  of  the  worm  must  be  such  that 

""  pitch  circumference  of  worm ; 

.'.  pitch  circumference  of  worm= —  —=6.87  inches. 

0.364 

6.87 
Pitch  diameter  of  worm  = — —  =2.2  inches. 

7T 

Pitch  diameter  of  wheel  =  —  =15.88  inches. 

Actual  distance  between  shafts  =  — —        —=9.04  inches. 


The  question  now  arises  whether  2.2  inches  is  a  great  enough 
pitch  diameter  for  the  worm.  If  the  thread  is  single  the  pitch 
=  2.5  inches  and  the  corresponding  dedendum=o.92  inch. 

Twice  this  dedendum  =  i.84  inches,  which  subtracted  from 
2.2  inches  would  only  leave  a  central  solid  core  of  0.36  inch  diam- 
eter for  the  worm.  It  is  obvious  without  computation  that  this 
would  not  sustain  the  torsional  moment.  If  the  double  thread 
were  used  the  central  core  would  have  a  diameter  of  1.28  inches. 

For  each  revolution  of  the  worm  the  length  of  the  path  of 
the  point  of  contact  or  the  distance  rubbed  over  equals  the  helix 
length  on  the  pitch  line.  This  is  the  hypothenuse  of  a  right- 
angle  triangle  whose  base  =6.87  inches  and  whose  altitude  = 
2.5  inches,  or  7.3  inches. 

At  300  revolutions  per  minute  the  distance  rubbed  through  in 

feet  per  minute  =  300  X  -1— =  182  feet.     At  182  feet  the  allowable 

pressure,  W,  between  the  teeth  may  equal  3500  Ibs.,  assuming 
bath  lubrication,  a  steel  worm,  and  a  bronze  wheel.    This  is  the 


TOOTHED  WHEELS  OR  GE4RS.  317 

pressure  applied  at  the  circumference  of  the  woim-wheel  in  the 
direction  of  the  axis  of  the  worm.  The  total  work  done  on  the 
worm-wheel  in  foot-pounds  per  minute  will  equal  W  multiplied 
by  the  pitch  velocity  of  the  wheel  in  feet  per  minute. 

This  wheel  makes  W=I5  revolutions  per  minute  and  its 
pitch  circumference  =f I  feet,  hence  its  pitch  velocity  =  15  XH 
=  62.5  feet  per  minute. 

3500X62.5  =218,750  ft.-lbs.  =6.63  H.P. 

This  same  amount  of  energy  is  transmitted  through  the  worm. 
The  twisting  moment  on  the  shaft  =Fr,  where  r  equals  the  pitch 
radius  of  the  worm.  .F=energy  transmitted  4- the  velocity  of 
the  point  of  application  of  the  force. 


6.87 


.Fr  =  1274X1.  i  =1401  in.-lbs. 

To  resist  this  there  is  a  circular  section  whose  strength  is 

t  xr,z 
represented  by  /8  —  -. 

fj  =  radius  of  core  of  worm  =0.64  inch; 
/,=unit  stress  in  outer  fiber; 
,      Fr      1401 


This  is  a  safe  value  for  steel.  Therefore  the  double-threaded 
worm  will  be  used  and  the  wheel  will  have  40  teeth  of  i^  inches 
circular  pitch. 


318  M  'A  'CHINE  DESIGN. 

For  the  strength  of  the  worm-wheel  teeth  Professor  Stribeck 
gives: 

W  =3$oPb  for  cast-iron  wheels; 
W  =  5$oPb  for  bronze  wheels; 
W  =  pressure  in  direction  of  worm  axis  in  pounds; 
P=  circular  pitch  in  inches; 
6=arc  EF,  Fig.  231,  in  inches. 

It  will  be  impossible  to  make  b  as  great  as  this  would  call  for 
in  the  problem  if  ^  =  3500  and  P  =  i|  inches. 

Had  the  distance  between  the  shafts  been  fixed  at  10  inches 
the  helix  angle  could  not  have  been  assumed  but  must  have  been 
calculated. 

The  pitch  radius  of  worm  would  have  been 


Pitch  circumference  =12.69  inches.  • 

2.  5 

Tangent  of  helix  angle  =  —  ^—  =0.1955. 


When  the  worm  and  worm-wheel  are  determined,  a  working 
drawing  may  be  made  as  follows:  Draw  AB,  Fig.  231,  the  axis 
of  the  worm-wheel,  and  locate  O,  the  projection  of  the  axis  of 
the  worm,  and  P,  the  pitch  point.  With  O  as  a  center  draw  the 
pitch,  full  depth,  and  addendum  circles,  G,  H,  and  K\  also 
the  arcs  CD  and  EF,  bounding  the  tops  of  the  teeth  and  the 
bottoms  of  the  spaces  of  the  worm-wheel.  Make  the  angle 
/3  =  9O°.  Below  EF  lay  off  a  proper  thickness  of  metal  to  support 
the  teeth  and  join  this  by  the  web  LM  to  the  hub  N.  The 
tooth  outlines  in  the  other  sectional  view  are  drawn  exactly 
as  for  an  involute  rack  and  pinion.  Full  views  might  be  drawn, 


TOOTHED    WHEELS   OR   GEARS. 


3*9 


but  they  involve  difficulties  of  construction,  and  do  not  give  any 
additional  information  to  the  workman. 

210.  Compound  Spur-gear  Chains. — Spur-gear  chains  may 
be  compound,  i.e.,  they  may  contain  links  which  carry  more  than 
two  elements.  Thus  in  Fig.  234  the  links  a  and  d  each  carry 
three  elements.  In  the  latter  case  the  teeth  of  d  must  be  counted 
as  two  elements,  because  by  means  of  them  d  is  paired  with  both 
b  and  c.  In  the  case  of  the  three -link  spur-gear  chain  the  wheels 
b  and  c  are  meshed  with  each  other,  and  a  point  in  the  pitch  circle 


FIG.  231 

of  c  moved  with  the  same  linear  velocity  as  a  point  in  the  pitch 
circle  of  b,  but  in.  the  opposite  direction.  In  Fig.  232  points  in 
all  the  pitch  circles  have  the  same  linear  velocity,  since  the  motion 
is  equivalent  to  rolling  together  of  the  pitch  circles  without  slip- 
ping; but  c  and  b  now  rotate  in  the  same  direction.  Hence  it 
is  seen  that  the  introduction  of  the  wheel  d  has  reversed  the 
direction  of  rotation,  without  changing  the  velocity  ratio.  The 


320  MACHINE  DESIGN. 

size  of  the  wheel,  d,  which  is  called  an  "idler,"  has  no  effect  upon 
the  motion  of  c  and  b.  It  simply  receives  upon  its  pitch  circle 
a  certain  linear  velocity  from  c,  and  transmits  it  unchanged 
to  b.  Hence  the  insertion  of  any  number  of  idlers  does  not 
affect  the  velocity  ratio  of  c  to  b,  but  each  added  idler  reverses 
the  direction  of  the  motion.  Thus,  with  an  odd  number  of 
idlers,  c  and  b  will  rotate  in  the  same  direction;  and  with  an 
even  number  of  idlers  c  and  b  will  rotate  in  opposite  directions. 

If  parallel  lines  be  drawn  through  the  centers  of  rotation  of 
a  pair  of  gears,  and  if  distances  be  laid  off  from  the  centers  on 
these  lines  inversely  proportional  to  the  angular  velocities  of  the 
gears,  then  a  line  joining  the  points  so  determined  will  cut  the 
line  of  centers  in  a  point  which  is  the  centre  of  the  gears.  In  Fig. 
232,  since  the  rotation  is  in  the  same  direction,  the  lines  have  to 


FIG.  232.  FIG.  233. 

be  laid  off  on  the  same  side  of  the  line  of  centers.  The  pitch 
radii  are  inversely  proportional  to  the  angular  velocities  of  the 
gears,  and  hence  it  is  only  necessary  to  draw  a  tangent  to  the  pitch 
circles  of  b  and  c,  and  the  intersection  of  this  line  with  the  line  of 
centers  is  the  centre,  be,  of  c  and  b.  The  centrodes  of  c  and  b  are 
c\  and  &i,  circles  through  the  point  be,  about  the  centers  of  c  and  b. 
Obviously  this  four-link  mechanism  may  be  replaced  by  a  three- 
link  mechanism,  in  which  GI  is  an  annular  wheel  meshing  with  a 
pinion  b\.  The  four  link  mechanism  is  more  compact,  however, 
and  usually  more  convenient  in  practice. 

The  other  principal  form  of    spur-gear  chain  is  shown  in 


TOOTHED   WHEELS   OR   GEARS.  321 

Fig.  233.  The  wheel  d  has  two  sets  of  teeth  of  different  pitch 
diameter,  one  pairing  with  c  and  the  other  with  b.  The  point 
bd  now  has  a  different  linear  velocity  from  cd,  greater  or  less  in 
proportion  to  the  ratio  of  the  radii  of  those  points.  The  angular 
velocity  ratio  may  be  obtained  as  follows: 

angular  velocity  d  _  C  .  .  .  cd 
angular  velocity  c      D  .  .  .  cd' 

ansrular  velocity  b      D  .  .  .  bd 
also, 


Multiplying, 


angular  velocity  d      B  .  .  .  bd' 

angular  velocity  b      C  .  .  .  cd  XD  .  .  .  bd 
angular  velocity  c     D  .  .  .  cdxB  .  .  .  bd' 


The  numerator  of  the  last  term  consists  of  the  product  of  the 
radii  of  the  "followers,"  and  the  denominator  consists  of  the 
product  of  the  radii  of  the  "drivers."  The  diameters  or  numbers 
of  teeth  could  be  substituted  for  the  radii. 

In  general,  the  angular  velocity  of  the  first  driver  is  to  the 
angular  velocity  of  the  last  follower  as  the  product  of  the  number 
of  teeth  of  the  followers  is  to  the  product  of  the  number  of  teeth 
of  the  drivers.  This  applies  equally  well  to  compound  spur-gear 
trains  that  have  more  than  three  axes.*  Therefore,  in  any  spur- 
gear  chain  the  velocity  ratio  equals  the  product  of  the  number  of 
teeth  in  the  followers  divided  by  the  product  of  the  number  of 
teeth  -in  the  drivers.  The  direction  of  rotation  is  reversed  if  the 
number  of  intermediate  axes  is  even,  and  is  not  reversed  if  the 
number  is  odd.  If  the  train  includes  annular  gears  their  axes 
would  be  omitted  from  the  number,  because  annular  gears  do 
not  reverse  the  direction  of  rotation. 

A  common  modification  of  the  chain  of  Fig.  233  is  shown  in 
Here  the  axis  of  the  gear  c  is  made  to  coincide  with 

*  Epicyclic  trains  excepted. 


322 


MACHINE  DESIGN. 


the  axis  of  b,  and  the  mechanism  is  known  as  a  reverted  gear  train, 
Probably  the  best  known  application  of  this  mechanism  is  that  of 
the  backgearing  of  the  ordinary  engine  lathe.  The  velocity  ratio 
of  c  and  b  is,  of  course,  not  altered  by  having  their  axes  coincide, 
and  it  is  equally  evident  that  one  of  them  only  may  be  keyed  to 
the  shaft  while  the  other  is  free  to  rotate  on  it. 


FIG.  233  A.  FIG.  233  B. 

2 10  a.  Epicyclic  Gearing. — In  the  gear  trains  of  the  preceding 
sections,  the  velocity  ratios  have  been  studied  with  reference  to 
a  fixed  member  to  which  each  gear  is  attached  by  a  turning  pair. 
Fig.  233  B  illustrates  such  a  simple  chain  of  three  links,  a,  b,  and 
c.  Considering  a  as  fixed  it  is  evident  that,  if  c  makes  m  turns 
per  minute  relatively  to  a,  causing  b  to  make  n  turns  per  minute 

relatively  to  a,  for  one  turn  of  c  relatively  to  a,  b  will  make     turns 

m 

relatively  to  a.     The  ratio  -  is  called  the  velocitv  ratio,  and    is 
m 

designated  by  r. 

If,  now,  one  of  the  gears,  c,  be  considered  as  the  fixed  link, 
and  it  is  desired  to  examine  the  action  of  the  mechanism  when 
a  is  swung  about  ac  as  center,  it  is  evident  that  a  different 
mechanism  is  obtained.  See  §8  and  §12.  The  action  can  be 
explained  under  the  general  laws  laid  down  in  these  sections 
but  can  be  understood  more  readily  by  reference  to  Fig.  233  C 
Such  mechanisms  are  known  as  epicydic  gear  trains,  because 
points  in  the  one  gear  describe  epicycloidal  curves  relatively  to 
the  other  gear.  The  name  has  no  connection  with  the  form  of 
the  gear  teeth  which  may  belong  to  the  cycloidal,  involute,  or 
any  other  system. 


TOOTHED   WHEELS  OR   GEARS. 


323 


Let  it  be  supposed  that  the  three  links  can  be  rigidly  locked 
together  and  while  so  held  are  given  a  complete  turn  about  the 
axis  ac,  in  a  clockwise  direction.  Owing  to  this,  b  will  make 
one  turn  in  a  clockwise  direction  about  its  own  axis  ab.  In 
position  i  the  arrow  is  seen  to  be  horizontal,  and  to  the  left  of  ab, 
at  2  it  is  vertical  and  above  ab,  at  3  horizontal  and  to  the  right, 
at  4  vertical  and  below,  and  at  i,  when  the  turn  about  ac  has 
been  completed,  it  is  once  more  horizontal  and  to  the  left  of  ab. 


Yiz.  033 C. 

The  arrow  on  b  has,  therefore,  made  a  complete  turn  about  ab 
as  axis,  and  if  one  line  of  the  rigid  body  b  has  made  such  a  turn 
the  whole  body  b  has  done  so.  But  in  swinging  the  locked 
mechanism  about  ac,  the  link  c  has  been  given  a  complete  revo- 
lution in  a  clockwise  direction.  This  is  contrary  to  the  original 
assumption  that  c  be  the  fixed  link,  i.e.,  remain  at  rest.  If, 
now,  the  mechanism  be  unlocked  and  c  be  given  a  complete 
revolution  in  a  counter-clockwise  direction  while  a  is  held  sta- 
tionary, the  result  will  be  the  same  as  though  c  had  not  been 
allowed  to  move  at  all.  But  this  counter-clockwise  revolution 
of  c  will  cause  b  to-  have  a  further  clockwise  rotation  about  its 

axis  of  —  =  r  turns.     The  total  number  of  turns  which  b  makes 

m 

about  its  axis  while  a  makes  one  turn  about  ac  will,  therefore, 
equal  i  +  r. 


324 


MACHINE  DESIGN. 


Had  an  idler  been  placed  between  b  and  c,  the  result  would 
have  been  to  cause  b  to  be  given  r  turns  in  a  counter-clockwise 
or  negative  direction,  when  c  was  brought  back  to  its  original 
position  and,  consequently,  b  would  make  i— r  revolutions  about 
its  axis  for  one  revolution  of  a  about  ac.  This  can  be  seen 
clearly  in  Fig.  233  D,  where  b  and  c  are  purposely  made  the 


FIG. 

same  size  so  that  r  =i  and,  hence,  i—r=o.  In  other  words, 
in  this  special  case  the  gear  b  does  not  rotate  about  its  axis 
at  all;  its  motion,  as  can  be  seen  from  positions  i,  2,  3,  and  4, 
being  merely  translation,  as  the  arrow  on  b  remains  always  parallel 
to  its  original  position. 

A  second  intermediate  gear,  or  idler,  would  again  reverse 
the  direction  of  &'s  motion,  making  the  revolutions  of  b  =i+r. 

The  general  law  may  be  stated  as  follows:  — "The  number  of 
revolutions  made  by  the  last  wheel  of  an  epicyclic  train  for  each 
revolution  of  the  arm  is  equal  to  the  one  plus  the  velocity  ratio 


TOOTHED   WHEELS  OR   GEARS.  325 

of  the  train  if  the  number  of  axes  in  the  train  be  even,  and  one 
minus  the  velocity  ratio  of  the  train  if  the  number  of  the  axes  be 
odd.  In  the  former  case  the  wheel  turns  in  the  same  sense  as 
the  arm ;  in  the  latter  in  the  opposite  sense,  unless  the  ratio  r 
is  less  than  unity."  (Kennedy — Mechanics  of  Machinery.) 

The  same  holds  if  there  are  no  annular  gears  in  the  train  or 
if  there  are  an  even  number  of  them.  If,  however,  there  be  one 
or  any  other  odd  number  of  annular  gears  in  the  train,  the  effect 
will  be  to  transpose  the  plus  and  minus  as  well  as  the  sense  of 
rotation. 

If  the  first  wheel  of  any  epicyclic  train  has  its  axis  fixed,  but 
has  itself  a  motion  of  rotation  about  this  axis  so  that,  for  example, 
it  makes  k  revolutions  for  each  revolution  of  the  arm,  then  the 
last  wheel  of  the  train  will  make  i±r±kr  revolutions  instead 
of  i±r.  The  sign  of  r  is  determined  as  before  but  the  sign  of 
kr  is  plus,  if  the  rotation  of  the  first  wheel  causes  the  last  wheel 
to  rotate  in  the  same  sense  as  the  arm,  and  minus,  if  the  rotation 
of  the  first  wheel  causes  the  last  wheel  to  rotate  in  a  sense  opposite 
that  of  the  arm. 

The  only  case  which  requires  special  attention  for  fear  of 


FIG.  233  E. 

incorrectly  determining  the  number  of  axes  is  where  the  gear 
train  of  Fig.  233,  which  has  three  axes,  is  given  the  reverted  form 
shown  in  Fig.  233  E,  which  apparently  has  but  two  axes.  For 
proper  analysis  it  is  necessary  to  consider  the  reverted  train 
the  same  as  the  original  form,  i.e.,  a  double  axis  is  counted  as 
two  single  ones  in  computing  the  number  of  axes  in  the  train. 


3-6  MACHINE  DESIGN. 

Problem. — Find  the  number  of  revolutions  c  will  make  about 
its  axis  for  each  revolution  of  the  arm  a\  d  being  considered  as 
the  fixed  link. 

d  has  101  teeth  and  meshes  with  b  which  has  100  teeth,  b'  is 
keyed  to  same  shaft  as  &,  has  99  teeth,  and  meshes  with  c,  which 
has  100  teeth.  If  this  were  an  ordinary  reverted  gear  train  with 
a  as  fixed  link,  then,  remembering  that  the  angular  velocity  of 
the  last  follower  is  to  the  angular  velocity  of  the -first  driver  as 
the  product  of  the  number  of  teeth  of  the  drivers  is  to  the  product 
of  the  number  of  teeth  of  the  followers,  for  one  turn  of  d,  c  would 

make  — — ^°_  =  _9_9_99.  turns  in  the  same  sense.    This  is  r,  the 
100X100       10000 

velocity  ratio  of  the  train.  Considering  the  train  as  an  epicyc- 
lic  one  with  d  as  fixed  link,  there  are  three  axes  and  no  annular 
gears  and  the  rule  would  be  that  for  one  turn  of  a  in  a  clock- 
wise direction  c  would  make  i  —  r  turns  about  its  axis  in 

the  same  sense,  equal  to  i  —         "   =  • • 

IOOOO         IOOOO 


CHAPTER  XVIII. 

SPRINGS. 

211.  Springs  Defined. — Usually  machine  members  are  required 
to  sustain  the  applied  forces  without  appreciable  yielding  and  are 
designed  accordingly;    but  certain  machine  members  are  useful 
because  of  considerable  yielding.    They  are  generally  called  springs. 

212.  Illustrations. — (a)  The  spring  of  a  safety-valve  on  a  steam- 
boiler  holds  the  valve  down  until  the  steam-pressure  reaches  the 
maximum  allowable  value;    then  it  yields  and  allows  steam  to 
escape  until  the  pressure  is  reduced,  when  it  closes  the  valve. 

(b)  The  springs  upon  which  a  locomotive-engine  is  supported 
prevent  the  transmission  of  the  full  effect  of  the  shocks,  due  to 
running,  to  the  working  parts  of  the  engine,  thereby  reducing  the 
resulting   stresses.      Car-springs    in    a    similar   manner    protect 
passengers  and  freight. 

(c)  "Bumper"  springs  reduce  stresses  in  cars  and  their  con- 
tents due  to  axial  shocks. 

(d)  The  springs  in  certain  steam-engine  governors  yield  under 
the  increased  centrifugal  force  of  the  governor  weights,  due  to 
increased  rotative  speeds,  and  allow  the  adjustment  of  the  valve- 
gear  to  the  changes  of  effort  and  load. 

(e)  Heavy  reciprocating  parts  are  often  brought  to  rest  without 
shock  and  are  then  helped  to  start  on  their  return  travel  by  the 
expanding  spring. 

(/)  A  power-hammer  strikes  a  "cushioned  blow"  because  of  the 

action  of  a  spring.    This  spring  may  be  of  steel,  rubber,  or  steam. 

(g)  Belt  connections  are  really  yielding  members  and  tend 

327 


328 


MACHINE  UESIGN. 


to  reduce  shocks  transmitted  through  them;  while  gears  (except 
rawhide  or  "hard-fiber  ")  yield  almost  imperceptibly  and  trans- 
mit shocks  almost  unchanged. 

(h)  Long  bolts  may  become  springs  for  the  reduction  of 
stress  due  to  shock. 

(i)  Springs  may  serve  for  the  storing  of  energy  which  is 
given  out  slowly  to  actuate  light-running  mechanisms,  like  clocks. 

213.  Cantilever  Springs. — Many  springs  are  simple  cantilevers 
with  end  loads.  (See  Fig.  234.) 


FIG.  234. 


FIG.  235. 


The  rectangular  spring  of  constant  width  b  and  height  (or 
thickness)  h,  with  a  load  F  applied  at  a  distance  /  from  the  sup- 
port gives,  from  the  laws  of  beams, 


}bh2 
F=6T 


and     ^ 


/  is  the  unit  stress  in  the  outer  fiber  in  pounds  per  square  inch, 
all  forces  being  expressed  in  pounds  and  all  dimensions  in  inches; 
J  is  the  total  deflection  in  inches  due  to  the  application  of  F; 
E  is  the  modulus  of  elasticity  of  the  material  used. 

For  a  flat  spring  of  uniform  breadth  &,  rectangular  cross- 
section,  top  surface  flat  and  lower  surface  a  parabola  in  outline, 
such  as  is  shown  in  Fig.  235, 


and     A 


6FP 
Ebhy 


SPRINGS. 


329 


The  same  equations  hold  approximately  for  the  cantilever 
spring  shown  in  Fig.  236.  They  also  hold  for  the  triangular 
spring  of  constant  depth  h  shown  in  Fig.  237. 


FIG.  236. 


FIG.  237. 


In  all  of  these  cases  obviously  the  yielding  varies  inversely 
as  h3,  and  the  strength  directly  as  h2;  hence,  if  h  be  increased 
to  obtain  required  strength,  the  yielding  will  be  decreased  as  the 
cube  of  h  while  the  strength  is  increased  only  as  the  square  of  h. 
Much  of  the  requisite  yielding  is  therefore  sacrificed  if  the  strength 
is  obtained  by  increasing  h. 

Inspection  of  the  same  equations  shows  that  increasing  the 
breadth  b  to  obtain  the  required  strength  decreases  the  deflection 
in  the  same  proportion.  In  springs, 
therefore,  where  yielding  is  to  be  kept 
large,  it  is  better  to  gain  requisite 
strength  by  varying  b;  while  in  a  beam 
should  be  as  large  as  possible  because 
here  deflection  is  to  be  reduced  to  the 
smallest  value.  If  the  spring  is  to  be  of 
tool  steel,  hardened  and  tempered,  thin 
material  is  better  suited  to  the  operation 
of  hardening. 

As    b  is    increased  with    a    constant 

small  value  of  h,  it  may  become  too  great  for  the  available 
space.  This  difficulty  is  overcome  as  shown  by  reference  to 
Fig.  238. 

Suppose  ABC  is  a  triangular  spring  designed  for  certain  con- 


FIG.  238. 


330  MACHINE  DESIGN. 

ditions  of  load  and  yielding.  A B  is  an  inconvenient  width. 
Divide  AB  into  equal  parts,  say  six.  Conceive  the  portion 
GFHBi  cut  off  and  placed  in  the  position  B^LK^i,  and  simi- 
larly conceive  that  EDKAi  occupies  the  position  AiMK\E\t 
and  the  two  parts  are  rigidly  joined  along  the  line  KiE\.  Also 
conceive  the  portion  ADE  moved  to  A\D\E\,  and  BFG  moved 
>  BiDiEi,  and  that  they  are  rigidly  joined  along  the  line  D\E±. 

The  amount  of  material  is  unchanged.  The  bending  force  is 
applied  in  nearly  the  same  way  to  the  portions  whose  position  is 
changed.  The  leaf  spring  is  therefore  practically  equivalent 
to  the  triangular  spring  from  which  it  is  made. 

214.  Springs  for  Axial  Loads. — Many  springs  are  subjected 
to  axial  loads;  they  are  usually  helical  in  form  as  shown  in 
Fig.  239. 


FIG.  239. 

F  may  act  to  stretch  or  compress  the  spring. 
Consider  the  cross-section  of  the  rod  to  be  circular. 
Let  F  =  load  in  pounds; 

d=  diameter  of  rod  in  inches; 
a=mean  radius  of  coil  in  inches; 
JV=number  of  coils; 

/=  developed  length  of  spring  in  inches  =  2?ra N; 
/.=  allowable  unit  shearing  stress  in  outer  fiber  in  pounds 

per  square  inch ; 

£,=modulus  of  shearing  elasticity  =f£; 
J=  extension  or  compression  in  inches. 
Then  the  following  equations  may  be  developed : 


and 


SPRINGS.  331 

In  a  helical  spring  for  an  axial  load  using  a  rectangular  cross- 
section  of  wire  the  axial  height  of  wire  =  h  and  radial  breadth  =  b, 
the  equations  become 


^2 

'  ' 


It  will  take  one  and  a  half  times  as  much  material  to  make  a  spring 
of  this  type  as  it  would  to  make  a  round-wire  helical  spring  of 
equal  strength. 

215.  Springs  for  Torsional  Movements.  —  Many  springs  come 


FIG.  240. 

under  class  (*)  as  mentioned  above.  The  general  case  is  shown 
in  Fig.  240.  The  spring  is  a  spiral  of  flat  wire  with  an  axial 
height  b  and  radial  depth  h. 

F  =  turning  force  in  pounds  ; 

R  =  lever-arm  in  inches  ; 

/  =  unit  stress  in  outer  fiber,  pounds  per  square  inch  ; 

/=  developed  length  of  spring; 

£  =  angular  deflection  ; 

A=  distance  in  inches  moved  through  by  the  point  of  appli- 
cation of  F. 

THen  ,_          „,  ,. 


- 

For  further  information   the   reader  is   referred   to   Reuleaux's 
"Constructor"  and  Trans.  A.  S.  M.  E.,  Vols.  V  and  XVI. 

-  -/-/£// 


CHAPTER  XIX. 

MACHINE   SUPPORTS. 

216.  General  Laws  for  Machine    Supports. — The  single-box 
pillar  support  is  best  and  simplest  for  machines  whose  size  and 
form  admit  of  its  use.     When  a  support  is  a  single  continuous 
member,  its  design  should  be  governed  by  the  following  principles : 

I.  The  amount  of  material  in  the  cross-section  is  determined 
by  the  intensity  of  the  load.     If  vibrations  are  also  to  be  sus- 
tained, the  amount  of  material  must  be  increased  for  this  purpose. 

II.  The  vertical  center  line  of  the  support  should  coincide 
with  the  vertical  line  through  the  center  of  gravity  of  the  part 
supported. 

III.  The  vertical  outlines  of  the  support  should  taper  slightly 
and  uniformly  on  all  sides.     If  they  were  parallel  they  would 
appear  nearer  together  at  the  bottom. 

IV.  The  external  dimensions  of  the  support  must  be  such 
that  the  machine  has  the  appearance  of  being  in  stable  equilibrium. 
The  outline  of  all  heavy  members  of  the  machine  supported  must 
be  either  carried  without  break  to  the  foundation,  or  if  they 
overhang,  must  be  joined  to  the  support  by  means  of  parabolic 
outlines,  or  by  the  straight  lines  of  the  brace  form. 

217.  Illustration. — In  Fig.  241  the  first  three  principles  may 
be   fulfilled,   but   there  is   an  appearance   of  instability.     It  is 
because  the  outline  of  the  "housing"  overhangs.     It  should  be 
carried  to  the  foundation  without  break  in  the  continuity  of  the 
metal,  as  in  Fig.  242. 

332 


MACHINE  SUPPORTS. 


133 


218.  Divided  Supports. — When  the  support  is  divided  up  into 
several  parts,  modification  of  these  principles  becomes  necessary, 
as  the  divisions  require  separate  treatment.  This  question  may 


FIG.  241.  FIG.  242. 

be  illustrated  by  lathe  supports.  In  Fig.  243  are  shown  three 
forms  of  support  for  a  lathe,  seen  from  the  end.  For  stability 
the  base  needs  to  be  broader  than  the  bed.  In  A  the  width  of 
base  necessary  is  determined  and  the  outlines  are  straight  lines. 
The  unnecessary  material  is  cut  away  on  the  inside,  leaving  legs 
which  are  compression  members  cf  correct  form.  The  crocs- 
brace  is  left  to  check  any  tendency  to  buckle.  For  convenience 
to  the  workmen  it  is  desirable  to  narrow  this  support  somewhat. 


FIG   243. 

without  narrowing  the  base.  The  cross-brace  converts  the  single 
compression  member  into  two  compression  members.  It  is  allow- 
able to  give  these  different  angles  with  the  vertical.  This  is  done 
in  B  and  the  straight  lines  are  blended  into  each  other  by  a  curve. 
C  shows  a  common  incorrect  form  of  lathe  support,  the  compres- 
sion members  from  the  cross-brace  downward  being  curved. 
There  is  no  reason  for  this  curved  form.  It  is  less  capable  of 


334 


MACHINE  DESIGN. 


bearing  its  compressive  load  than  if  it  were  straight,  and  is  no 
more  stable  than  the  form  B,  the  width  of  base  being  the  same. 


FIG.  244. 

Consider  the  lathe  supports  from  the  front.  Four  forms  are 
shown  in  Fig.  244.  If  there  were  any  force  tending  to  move  the 
bed  of  the  lathe  endwise  the  forms  B  and  C  would  be  allowable. 
But  there  is  no  force  of  this  kind,  and  the  correct  form  is  the 
one  shown  in  D.  Carrying  the  foot  out  as  in  A,  B,  and  C  in- 
creases the  distance  between  supports  (the  bed  being  a  beam 
with  end  supports  and  the  load  between) ;  this  increases  the  de- 
flection and  the  fiber  stress  due  to  the  load.  This  increase  in 
stress  is  probably  not  of  any  serious  importance,  but  the  prin- 
ciple should  be  regarded  or  the  appearance  of  the  machine  will 
not  be  right.  If  the  supports  were  joined  by  a  cross-member, 


FIG    245. 

as  in  Fig.  245,  they  would  be  virtually  converted  into  a  single 
support,  and  should  then  taper  from  all  sides. 

219.  Three-point  Support. — If  a  machine  be  supported  on 
a  single -box  pillar,,  change  in  the  form  of  the  foundation  cannot 
induce  stress  in  the  machine  frame  tending  to  change  its  form. 
If,  however,  the  machine  is  supported  on  four  or  more  legs  the 


MACHINE  SUPPORTS.  335 

foundation  might  sink  away  from  one  or  more  of  them  and  leave 
a  part  unsupported.  This  might  cause  torsional  or  flexure  stress 
in  some  part  of  the  machine,  which  might  change  its  form  and 
interfere  with  the  accuracy  of  its  action. 

But  if  the  machine  is  supported  on  three  points  this  cannot 
occur,  because  if  the  foundation  should  sink  under  any  one  of 
the  supports  the  support  would  follow  and  the  machine  would 
still  rest  on  three  points.  When  it  is  possible,  therefore,  a  ma- 
chine which  cannnot  be  carried  on  a  single  pillar  should  be  sup- 
ported on  three  points.  Many  machines  are  too  large  for  three- 
point  support,  and  the  resource  is  to  make  the  bed,  or  part  sup- 
ported, of  box  section  and  so  rigid  that  even  if  some  of  the  legs 
should  be  left  without  foundation  the  part  supported  would  still 
maintain  its  form.  More  supports  are  often  used  than  are  necessary. 
Thus,  if  a  lathe  has  two  pairs  of  legs  like  those  shown  in  B, 
Fig.  243,  and  these  are  bolted  firmly  to  the  bed,  there  will  be 
four  points  of  support.  But  if,  as  suggested  by  Professor  Sweet, 
one  of  these  pairs  be  connected  to  the  bed  by  a  pin  so  that  the 
support  and  the  bed  are  free  to  move  relatively  to  each  other 
about  the  pin,  as  in  Fig.  246,  then  this  is  equivalent  to  a  single 
support,  and  the  bed  will  have  three  points  of  support,  and  will 
maintain  its  form  independently  of  any  change  in  the  foundation. 
This  is  of  special  importance  when  the  machines  are  to  be  placed 
upon  yielding  floors. 

220.  Reducing  Number  of  Supports. — Fig.  247  shows  another 


FIG.  246.  FIG.  247. 

case  in  which  the  number  of  supports  may  be  reduced  without 
sacrifice.     In  A  three  pairs  of  legs  are  used.     There  are  therefore 


336 


MACHINE  DESIGN. 


six  points  of  support.  In  B  two  pairs  of  legs  are  used  and  one 
may  be  connected  by  a  pin,  and  there  will  be  but  three  points  of 
support.  The  chance  of  the  bed  being  strained  from  changing 
foundation  has  been  reduced  from  6  in  A  to  o  in  B.  The  total 
length  of  bed  is  12  feet,  and  the  unsupported  length  is  6  feet 
in  both  cases. 

221.  Further   Correct   Methods. — Figs.    248   and    249    show 
correct  methods  of  support  for  small  lathes  and  planers,  due  to 


FIG.  248. 


FIG.  249. 


Professor  Sweet.  In  Fig.  248  the  lathe  "head-stock"  has  its 
outlines  carried  to  the  foundation  by  the  box  pillar;  a  represents 
a  pair  of  legs  connected  to  the  bed  by  a  pin  connection,  and 
instead  of  being  placed  at  the  end  of  the  bed  it  is  moved  in  some- 
what, the  end  of  the  bed  being  carried  down  to  the  support  by  a 
parabolic  outline.  The  unsupported  length  of  bed  is  thereby 
decreased,  the  stress  on  the  bed  is  less,  and  the  bed  will  maintain 
its  form  regardless  of  any  yielding  of  the  floor  or  foundation. 
In  Fig.  249  the  housings,  instead  of  resting  on  the  bed  as  is  usual 
in  small  planers,  are  carried  to  the  foundation,  forming  two  of 
the  sifpports;  the  other  is  at  a  and  has  a  pin  connection  with  the 
bed,  which  being  thus  supported  on  three  points  cannot  be  twisted 
or  flexed  by  a  yielding  foundation. 


CHAPTER  XX. 

MACHINE    FRAMES. 

222.  Open-side  Frame. — Fig.  250  shows  an  open-side  frame, 
such  as  is  used  for  punching  and  shearing  machines.  During  the 
action  of  the  punch  or  shear  a  force  is  applied  to  the  frame  tending 
to  separate  the  jaws.  This  force  may  be  represented  in  magnitude, 
direction,  and  line  of  action  by  P.  It  is  required  to  find  the 
resulting  stresses  in  the  three  sections  AB,  CD,  and  EF.  Con- 
sider AB.  Let  the  portion  above  this  section  be  taken  as  a  free 
body.  The  force  P,  Fig.  251,  and  the  opposing  resistances  to 


FIG.  250. 

deformation  of  the  material  at  the  section  AB,  are  in  equilibrium. 
Let  H  be  the  projection  of  the  gravity  axis  of  the  section  AB, 
perpendicular  to  the  paper.  Two  equal  and  opposite  forces,  PI 
and  PZ,  may  be  applied  at  H  without  disturbing  the  equilibrium. 
Let  PI  and  P^  be  each  equal  to  P,  and  let  their  line  of  action  be 
parallel  to  that  of  P.  The  free  body  is  now  subjected  to  the 
action  of  an  external  couple,  PI,  and  an  external  force,  PI.  The 

337 


338 


MACHINE  DESIGN. 


couple  produces  flexure  about  H,  and  the  force  PI  produces  tensile 
stress  in  the  section  AB.  The  flexure  results  in  a  tensile  stress 
varying  from  a  maximum  value  in  the  outer  fiber  at  A  to  zero 
at  H,  and  a  compressive  stress  varying  from  a  maximum  in  the 
outer  fiber  at  B  to  zero  at  H .  This  may  be  shown  graphically 
at  JK.  The  ordinates  of  the  line  LM  represent  the  varying 
stress  due  to  flexure;  while  ordinates  between  LM  and  NO 
represent  the  uniform  tensile  stress.  This  latter  diminishes  the 
compressive  stress  at  B,  and  increases  the  tensile  stress  at  A. 
The  tensile  stress  per  square  inch  at  A  therefore  equals  /+/i; 
where  /  equals  the  unit  fiber  stress  due  to  flexure  at  ^4,and  /i 


FIG.  252. 


Pic  P 

equals  the  unit  tensile  stress  due  to  PI.     Now  /=—=-,  and  /i  =-7-; 

in  which  c  =the  distance  from  the  gravity  axis  to  the  outer  fiber  = 
AH,  and  7  =  the  moment  of  inertia  of  the  section  about  H,  and 
A  =area  of  the  cross-section  AB. 

223.  Problem. — Let  it  be  required  to  design  the  frame  of  a 
machine  to  punch  f-inch  holes  in  J-inch  steel  plates,  18  inches 
from  the  edge.  The  surface  resisting  the  shearing  action  of  the 
punch  =  7rXf"X|"=i.i8  square  inch.  The  ultimate  shearing 
strength  of  the  material  is  say  50,000  Ibs.  per  square  inch.  The 
total  force  P,  which  must  be  resisted  by  the  punch  frame  =  50,000 
X 1. 18=59,000  Ibs. 


MACHINE  FRAMES.  339 

The  material  and  form  for  the  frame  must  first  be  selected. 
The  form  is  such  that  forged  material  is  excluded,  and  difficulties 
of  casting  and  high  cost  exclude  steel  casting.  The  material, 
therefore,  must  be  cast  iron.  Often  the  same  pattern  is  used 
both  for  the  frame  of  a  punch  and  shear.  In  the  latter  case, 
when  the  shear  blade  begins  and  ends  its  cut,  the  force  is  not 
applied  in  the  middle  plane  of  the  frame,  but  considerably  to  one 
side,  and  a  torsional  stress  results  in  the  frame.  Combined  torsion 
and  flexure  are  best  resisted  by  members  of  box  form.  The  frame 
will  therefore  be  made  of  cast  iron  and  of  box  section.  The  dimen- 
sion AB  may  be  assumed  so  that  its  proportion  to  the  "reach" 
of  the  punch  appears  right;  the  width  and  thickness  of  the  cross- 
section  may  also  be  assumed.  From  these  data  the  maximum 
stress  in  the  outer  fiber  may  be  determined.  If  this  is  a  safe 
value  for  the  material  used  the  design  will  be  right. 

Let  the  assumed  dimensions  be  as  shown  in  Fig.  252.    Then 

A  =  b\d\  —b2d2='j8  square  inches. 


12 

=  3002  bi-quadratic  inches. 

c=di-^2=9";  I  =  the  reach  of  the  punch  +^=27";  P=  59,000  Ibs., 
as  determined  above.     Then 


Pic     59000  X  27  X  9 

™    "^r-  '=4776' 


/i  +/  =  5532  =maximum  stress  in  the  section. 

The  average  strength  of  cast  iron  such  as  is  used  for  machinery 
castings,  is  about  20,000  Ibs.  per  square  inch.  The  factor  of 
safety  in  the  case  assumed  equals  20,000-^5532=3.65.  This  is 
too  small.  There  are  two  reasons  why  a  large  factor  of  safety 


34° 


MACHINE  DESIGN. 


should  be  used  in  this  design:  I.  When  the  punch  goes  through 
the  plate  the  yielding  is  sudden  and  a  severe  stress  results.  This 
stress  has  to  be  sustained  by  the  frame,  which  for  other  reasons 
is  made  of  unresilient  material.  II.  Since  the  frame  is  of  cast 
iron,  there  will  necessarily  be  shrinkage  stresses  which  the  frame 
must  sustain  in  addition  to  the  stress  due  to  external  force's. 
These  shrinkage  stresses  cannot  be  calculated  and  therefore  can 
only  be  provided  against  by  a  large  factor  of  safety. 

Cast  iron  is  strong  to  resist  compression  and  weak  to  resist 
tension,  and  the  maximum  fiber  stress  is  tension  on  the  inner 
side.  The  metal  can  therefore  be  more  satisfactorily  distributed 
than  in  the  assumed  section,  by  being  thickened  where  it  sustains 
tension,  as  at  A,  Fig.  253.  If,  however,  there  is  a  very  thick 
body  of  metal  at  a,  sponginess  and  excessive  shrinkage  would 
result.  The  form  B  would  be  better,  the  metal  being  arranged 
for  proper  cooling  and  for  the  resisting  of  flexure  stress. 


FIG.  253. 


FIG.  254. 


Dimensions  may  be  assigned  to  a  section  like  B  and  the 
cross-section  may  be  checked  for  strength  as  before.  See  Fig.  254. 
GG,  a  line  through  the  center  of  gravity  of  the  section,  is  found  to 
be  at  a  distance  of  7.05  inches  from  the  tension  side.*  The 

*  A  simple  and  satisfactory  method  for  obtaining  a  close  approximation  to 


MACHINE  FRAMES.  34  1 

required  values  are  as  follows:  £  =  7.05  inches;  /  =  reach  of 
punch  +c  =  18  +  7.05  =25.05  inches;  ^=156.25  square  inches; 
/  =  5032.5  bi-quadratic  inches  ;  P  =  59,000  Ibs. 


Then  /i=     =  =377.61bs.; 

A     156.25     ' 

Pic     59000X25.05X7.05 
}  =  —r  =  —  —  -  =  2070.4  Ibs. 

I  5032-5 

/i+/  =  2448  Ibs.  =  maximum  fiber  stress  in  the  section.  The 
factor  of  safety  =20,000^2448  =8.17.*  This  section,  therefore, 
fulfills  the  requirement  for  strength,  and  the  material  is  well 
arranged  for  cooling  with  little  shrinkage,  and  without  spongy 
spots.  The  gravity  axis  may  be  located,  and  the  value  of  /  deter- 
mined by  graphic  methods.  See  Hoskins's  "Graphic  Statics."  f 
Let  the  section  CD,  Fig.  250,  be  considered.  Fig.  255  shows 
the  part  at  the  left  of  CD  free.  K  is  the  projection  of  the  gravity 
axis  of  the  section.  As  before,  put  in  two  opposite  forces,  Pa 
and  P4,  equal  to  each  other  and  to  P,  and  having  their  common 
line  of  action  parallel  to  that  of  P,  at  a  distance  l\  from  it.  P 
and  P<inow  form  a  couple,  whose  moment  =  P/i,  tending  to  produce 
flexure  about  K.  P3  must  be  resolved  into  two  components, 
one  P3J,  at  right  angles  to  the  section  considered,  tending  to 
produce  tensile  stress;  and  the  other  JK,  parallel  to  the  section, 
tending  to  produce  shearing  stress.  The  greatest  unit  tensile 

the  true  gravity  axis  of  irregular  figures  is  as  follows:  On  a  piece  of  thin  but  unt- 
form  cardboard  lay  out  the  figure  to  scale.  Cut  it  out  carefully  with  a  sharp 
knife.  Balance  the  figure  exactly,  by  trial,  on  a  knife-edge.  The  line  of  contact 
with  the  knife-edge  is  the  gravity  axis.  Its  position  may  be  marked  and  its  loca- 
tion measured  to  scale. 

*  This  discussion  neglects  the  action  of  gravity  which  would  exert  a  counter- 
balancing moment,  reducing  the  maximum  tensile  fiber  stress  below  the  value 
found.  This  makes  the  actual  factor  of  safety  greater  than  the  apparent  facto? 
of  safety. 

t  The  student  will  be  familiar  with  analytical  methods  for  their  determina 
tion  from  his  study  of  the  ''Mechanics  of  Materials." 


342 


MACHINE  DESIGN. 


stress  in  this  section  will  equal  the  sum  of  that  due  to  flexure 
and  that  due  to  tension 

...      Plic    PZJ 

-/+/!-— +-f-- 

JK 

The  greatest  unit  shear  =/s==^- 

In  the  section  FE,  Fig.  250,  which  is  parallel  to  the  line  of 
faction  of  P,  equal  and  opposite  forces,  each  =  P,  may  be  intro- 
duced, as  PS  and  PQ.  P  and  P6  will  then  form  a  couple  with  an 


FIG.  256. 

arm  12,  and  P5  will  be  wholly  applied  to  produce  shearing  stress. 
The  maximum  unit  tensile  stress  in  this  section  will  be  that  due 
to  flexure,  /=P/2c-^/,  and  the  maximum  unit  shear  will  be 
}8  =  P+A.  Any  section  may  be  thus  checked. 


MACHINE  FRAMES. 


343 


The  dimensions  of  several  sections  being  found,  the  outline 
curve  bounding  them  should  be  drawn  carefully,  to  give  good 
appearance.  The  necessary  modifications  of  the  frame  to  pro- 
vide for  support,  and  for  the  constrainment  of  the  actuating 
mechanism,  may  be  worked  out  as  in  Fig.  256.  A  is  the  pinion 
on  the  pulley  shaft  from  which  the  power  is  received;  B  is  the 
gear  on  the  main  shaft ;  C,  D,  and  G  are  parts  of  the  frame  added 
to  supply  bearings  for  the  shafts;  E  furnishes  the  guiding  surfaces 
for  the  punch  "slide."  The  method  of  supporting  the  frame  is 
shown,  the  support  being  cut  under  at  F  for  convenience  to  the 
workman.  The  parts  C,  D,  E,  and  G  can  only  be  located  after 
the  mechanism  train  has  been  designed. 


FIG.  257. 

224.  Slotting-machine  Frame. — See  Fig.  257.  It  is  specified 
that  the  slotter  shall  cut  at  a  certain  distance  from  the  edge  of 
any  piece,  and  the  dimension  AH  is  thus  determined.  The 
table  G  must  be  held  at  a  convenient  height  above  the  floor,  and 
RK  must  provide  for  the  required  range  of  "feed."  K  is  cut 
under  for  convenience  of  the  workman,  and  carried  to  the  floor 
line  as  shown.  It  is  required  to  "slot  "  a  piece  of  given  vertical 
dimension,  and  the  distance  from  the  surface  of  the  table  to  E 
is  thus  determined.  Let  the  dimension  LM  be  assumed  so  that 


344  MACHINE  DESIGN. 

it  shall  be  in  proper  proportion  to  the  necessary  length  and  height 
of  the  machine.  The  curves  LS  and  MT  may  be  drawn  for 
bounding  lines  of  a  box  frame  to  support  the  mechanism.  M 
should  be  carried  to  the  floor  line  as  shown,  and  not  cut  under. 
None  of  the  part  DNE,  nor  that  which  serves  to  support  the 
cone  and  gears  on  the  other  side  of  the  frame,  should  be  made 
flush  with  the  surface  LSTM,  because  nothing  should  interfere 
with  the  continuity  of  the  curves  LS  and  TM .  The  supporting 
frame  of  a  machine  should  be  clearly  outlined,  and  other  parts  should 
appear  as  attachments.  The  member  VW  should  be  designed  so 
that  its  inner  outline  is  nearly  parallel  to  the  outline  of  the  cone 
pulley,  and  should  be  joined  to  the  main  frame  by  a  curve.  The 
outer  outline  should  be  such  that  the  width  of  the  member  increases 
slightly  from  W  to  V,  and  should  also  be  joined  to  the  main 
frame  by  a  curved  outline.  In  any  cross-section  of  the  frame,, 
as  XX,  the  amount  of  metal  and  its  arrangement  may  be  con- 
trolled by  the  core.  It  is  dictated  by  the  maximum  force,  P,. 
which  the  tool  can  be  required  to  sustain.  The  tool  is  carried  by 
the  slider  of  a  slider-crank  chain.  Its  velocity  varies,  therefore,, 
from  a  maximum  near  mid-stroke,  to  zero  at  the  upper  and  lower 
ends  of  its  stroke.  The  belt  which  actuates  the  mechanism  runs 
on  one  side  of  the  steps  of  the  cone  pulley,  at  a  constant  velocity. 
Suppose  that  the  tool  is  set  (accidentally)  so  that  it  strikes  the 
table  just  before  the  slider  has  reached  the  lower  end  of  its  stroke. 
The  resistance,  R,  offered  by  the  tool  to  being  stopped,  multiplied 
by  its  (very  small)  velocity,  equals  the  difference  of  belt  tension 
multiplied  by  the  belt  velocity  (friction  and  inertia  neglected).* 
R,  therefore,  would  vary  inversely  as  the  slider  velocity,  and  hence 
may  be  very  great.  Its  maximum  value  is  indeterminate.  A 
"breaking  piece  "  may  be  put  in  between  the  tool  and  the  crank. 
Then  when  R  reaches  a  certain  value,  the  breaking  piece  fails. 

*  See  Chapter  V. 


MACHINE  FRAMES. 


345 


The  stress  in  the  stress-members  of  the  machine  is  thereby  limited 
to  a  certain  definite  value.  From  this  value  the  frame  may  be 
designed.  Let  P  =  up  ward  force  against  the  tool  when  the  break- 
ing piece  fails.*  Let  /  =  the  horizontal  distance  from  the  line  of 
action  P  to  the  gravity  axis  of  the  section  XX.  Then  the  section 
XX  sustains  flexure  stress  caused  by  the  moment  PI,  and  tensile 
^tre.s  equal  to  P.  The  maximum  unit  stress  in  the  section 

Pic      P 


A  section  may  be  assumed  and  checked  for  safety,  as  for  the 
punching-machine  in  §  223. 

225.  Stresses  in  the  Frame  of  a  Side-crank  Steam-engine.  — 
Fig.  258  is  a  sketch  in  plan  of  a  side-crank  engine  of  the  "girder 


FIG.  258. 

bed  ''  type.  The  supports  are  under  the  cylinder  C,  the  main 
bearing  E,  and  the  out-board  bearing  D.  A  force  P  is  applied, 
in  the  center  line  of  the  cylinder,  and  acts  alternately  toward  the 
right  and  toward  the  left.  In  the  first  case  it  tends  to  separate 
the  cylinder  and  main  shaft;  and  in  the  second  case  it  tends  to 
bring  them  nearer  together.  The  frame  resists  these  tendencies 
with  resulting  internal  stresses. 

Let  the  stresses  in  the  section  AB  be  considered.  The  end  of 
ths  frame  is  shown  enlarged  in  Fig.  259.  If  the  pressure  from  the 
piston  is  toward  the  right,  the  stresses  in  AB  will  be:  I.  Flexure 


P  is  limited  to  the  friction  due  to  screwing  up  the  four  bolts  which  hold  the 


tool. 


346 


MACHINE  DESIGN. 


due  to  the  moment  PI,  resulting  in  tensile  stress  below  the  gravi'ty 
axis,  N,  with  a  maximum  value  at  b,  and  a  compressive  stress 
above  N  with  a  maximum  value  at  a.  II.  A  direct  tensile  stress, 
=  P,  distributed  over  the  entire  section,  resulting  in  a  unit  stress  = 
P+A=}i  Ibs.  per  square  inch.  This  is  shown  graphically  at  n, 
Fig.  259.  ai&i  is  a  datum  line  whose  length  equals  ab.  Tensions 
are  laid  off  toward  the  right  and  compressions  toward  the  left. 


(n) 


FIG.  259. 


The  stress  due  to  flexure  varies  directly  as  the  distance  from  the 
neutral  axis  NI,  being  zero  at  NI.  If,  therefore,  b\,c\  represents 
the  tensile  stress  in  the  outer  fiber,  then  c\k\  drawn  through  N\ 
will  be  the  locus  of  the  ends  of  horizontal  lines,  drawn  through 
all  points  of  a\b\,  representing  the  intensity  of  stress  in  all  parts 
of  the  section,  due  to  flexure.  If  c\d\  represent  the  unit  stress 
due  to  direct  tension,  then,  since  this  is  the  same  in  all  parts  of 
the  section,  it  will  be  represented  by  the  horizontal  distance  be- 
tween the  parallel  lines  c\k\  and  d\e\.  This  uniform  tension 
increases  the  tension  bic\  due  to  flexure,  causing  it  to  become  b^d\  ; 
and  reduces  the  compression  k\ai,  causing  it  to  become  e^a\. 
The  maximum  stress  in  the  section  is  therefore  tensile  stress  in 
the  lower  outer  fiber,  and  is  equal  to  b\d\. 

When  the  force  P  is  reversed,  acting  toward  the  left,  the 
stresses  in  the  section  are  as  shown  at  m,  Fig.  259:  compression 
due  to  flexure  in  the  lower  outer  fiber  equal  to  c2b2;  tension  due 
to  flexure  in  the  upper  outer  fiber  equal  to  a2k2;  and  uniform 
compression  over  the  entire  surface  equal  to  d2c2.  This  latter 
increases  the  compression  in  the  lower  outer  fiber  from  b2c2  to 
b2d2,  and  decreases  the  tension  in  the  upper  outer  fiber  from  a2k2 


MACHINE  FRAMES.  347 

to  a2^2-  The  maximum  stress  in  the  section  is  therefore  compres- 
sion in  the  lower  outer  fiber  equal  to  b^.  The  maximum  stress, 
therefore,  is  always  in  the  side  of  the  frame  next  to  the  connecting- 
rod. 

If  the  gravity  axis  of  the  cross-section  be  moved  toward  the 
connecting-rod,  the  stress  in  the  upper  outer  fiber  will  be  increased, 
and  that  in  the  lower  outer  fiber  will  be  proportionately  decreased. 
The  gravity  axis  may  be  moved  toward  the  connecting-rod  by 
increasing  the  amount  of  material  in  the  lower  part  of  the  cross- 
section  and  decreasing  it  in  the  upper  part. 

The  stress  in  any  other  section  nearer  the  cylinder  will  be  due 
to  the  same  force,  P,  as  before;  but  the  moment  tending  to  pro- 
duce flexure  will  be  less,  because  the  lever  arm  of  the  moment  is 
less  and  the  force  constant. 

226.  Heavy-duty  Engine  Frame. — Suppose  the  engine  frame 
to  be  of  the  type  which  is  continuous  with  the  supporting  part  as 
shown  in  Fig.  260.  Let  Fig.  261  be  a  cross-section,  say  at  AB. 
O  is  the  center  of  the  cylinder.  The  force  P  is  applied  at  this 


FIG.  261. 


point  perpendicular  to  the  paper.  C  is  the  center  of  gravity  of 
the  section  (the  intersection  of  two  gravity  axes  perpendicular 
to  each  other,  found  graphically).  Join  C  and  O,  and  through 


348  MACHINE  DESIGN. 

C  draw  XX  perpendicular  to  CO.  Then  XX  is  the  gravity 
axis  about  which  flexure  will  occur.*  The  dangerous  stress  will 
be  at  F,  and  the  value  of  c  will  be  the  perpendicular  distance  from 
F  to  XX.  The  moment  of  inertia  of  the  cross-section  about 
XX  may  be  found,  =/;  /,  the  lever-arm  of  P,  =OC.  The 
stress  at  F,  /+/i  must  be  of  safe  value. 

/=— ,  in  known  terms. 


area  of  section 


in  known  terms. 


227.  Closed  Frames. — Fig.  262  shows  a  closed  frame.  The 
members  G  and  H  are  bolted  rigidly  to  a  cylinder  C  at  the  top,  and 
to  a  bedplate,  DD,  at  the  bottom.  A  force  P  may  act  in  the  center 
line,  either  to  separate  D  and  C,  or  to  bring  them  nearer  together. 
The  problem  is  to  design  G,  H,  and  D  for  strength.  If  the  three 
members  were  "pin  connected"  (see  Fig.  263),  the  reactions  of 
C  upon  A  and  B  at  the  pins  would  act  in  the  lines  EF  and  GH. 
Then  if  P  acts  to  bring  D  and  C  nearer  together,  compression 
results  in  A,  the  line  of  action  being  EF;  compression  results 
in  B,  the  line  of  action  being  GH.  These  compressions  being  in 
equilibrium  with  the  force  P,  their  magnitude  may  be  found  by 
the  triangle  of  forces.  From  these  values  A  and  B  may  be  designed, 
C  is  equivalent  to  a  beam  whose  length  is  /,  supported  at  both 

*  This  is  not  strictly  true.  If  OC  is  a  diameter  of  the  "ellipse  of  inertia," 
flexure  will  occur  about  its  conjugate  diameter.  If  the  section  of  the  engine  frame 
is  symmetrical  with  respect  to  a  vertical  axis,  OC  is  vertical,  and  its  conjugate 
diameter  XX  is  horizontal.  Flexure  would  occur  about  XX,  and  the  angle  be- 
tween OC  and  XX  would  equal  90°.  As  the  section  departs  from  symmetry 
about  a  vertical,  XX,  at  right  angles  to  OC,  departs  from  OC's  conjugate,  and 
hence  does  not  represent  the  axis  about  which  flexure  occurs.  In  sections  like 
Fig.  259,  the  error  from  making  ^  =  90°  is  unimportant.  When  the  departure 
from  symmetry  is  very  great,  however,  OC's  conjugate  should  be  located  and 
used  as  the  axis  about  which  flexure  occurs.  For  method  of  drawing  "  ellipse  of 
inertia"  see  Hoskins's  "  Graphic  Statics." 


MACHINE  FRAMES. 


349 


ends,  sustaining  a  transverse  load  P,  and  tension  equal  to  the 
horizontal  component  of  the  compression  in  A  or  B.  The  data 
for  its  design  would  therefore  be  available.  Reversing  the  direc- 
tion of  P  reverses  the  stresses;  the  compression  in  A  and  B  becomes 
tension;  the  flexure  moment  tends  to  bend  C  convex  downward 
instead  of  upward,  and  the  tension  in  C  becomes  compression. 
But  when  the  members  are  bolted  rigidly  together,  as  in 

FIG.  262. 


Fig.  262,  the  lines  of  the  reactions  are  indeterminate.  Assump- 
tions must  therefore  be  made.  Suppose  that  G  is  attached  to  D 
by  bolts  at  E  and  A.  Suppose  the  bolts  to  have  worked  slightly 
loose,  and  that  P  tends  to  bring  C  and  D  nearer  together.  There 
would  be  a  tendency,  if  the  frame  yields  at  all,  to  relieve  pressure 
at  E  and  to  concentrate  it  at  A.  The  line  of  the  reaction  would 
pass  through  A  and  might  be  assumed  to  be  perpendicular  to 


350  MACHINE  DESIGN. 

the  surface  AE.  Suppose  that  P  is  reversed  and  that  the  bolts 
at  A  are  loosened,  while  those  at  E  are  tight.  The  line  of  the 
reaction  would  pass  through  E,  and  might  be  assumed  to  be 
perpendicular  to  EA.  MN  is  therefore  the  assumed  line  of  the 
reaction,  and  the  intensity  R=P+2.  In  any  section  of  G,  as 
XX,  let  Kr  be  the  projection  of  the  gravity  axis.  Introduce  at 
K'  two  equal  and  opposite  forces  equal  to  R  and  with  their  lines 
of  action  parallel  to  that  of  R.  Then  in  the  section  there  is  flex- 
ure stress  due  to  the  flexure  moment  Rl,  and  tensile  stress  due  to 
the  component  of  Rz  perpendicular  to  the  section,  =^3.  Then 
the  maximum  stress  in  the  section  =/+/i. 


A  section  may  be  assumed,  and  A,  I,  and  c  become  known; 
the  maximum  stress  also  becomes  known,  and  may  be  compared 
with  the  ultimate  strength  of  the  material  used. 

Obviously  this  resulting  maximum  stress  is  greater  when  the 
line  of  the  reaction  is  MN  than  when  it  is  KL.  Also  it  is  greater 
when  MN  is  perpendicular  to  EA  than  if  it  were  inclined  more 
toward  the  center  line  of  the  frame.  The  assumptions  therefore 
give  safety.  If  the  force  P  could  only  act  downward,  as  in  a 
steam  hammer,  KL  would  be  used  as  the  line  of  the  reaction. 

The  part  D  in  the  bolted  frame  is  not  equivalent  to  a  beam 
with  end  supports  and  a  central  load  like  C,  Fig.  263,  but  more 
nearly  a  beam  built  in  at  the  ends  with  central  load,  and  it  may 
be  so  considered,  letting  the  length  of  the  beam  equal  the  hori- 

zontal distance  from  E  to  F,  =/i.    Then  the  stress  in  the  mid- 

pi 

section  will  be  due  to  the  flexure  moment  ~,  and  the  maximum 

o 

stress  =  /  =  ~nj~'    The  values  c  and  I  may  be  found  for  an  assumed 
section,  and  /  becomes  known. 


MACHINE  FRAMES.  35 J 

228.  Steam-hammer  Frames. — Steam-hammers  are  made  both 
with  "open-side  "  and  "closed  "  frames.     They  may  therefore  be 
designed  by  methods  already  given,  if  the  maximum  force  applied 
is  known.     The  problem  is,  therefore,  to  find  the  value  of  this 
maximum  force. 

There  are  two  types  of  steam-hammers: 

Type  i.  Single-acting.  A  heavy  hammer-head  attached  to 
a  steam-piston  is  raised  to  a  certain  height  by  steam  admitted 
under  the  piston.  The  steam  is  then  exhausted  and  the  hammer- 
head with  attached  parts  falls  by  gravity  to  its  original  position. 
The  energy  of  the  blow  =  Wl,  where  W  is  the  falling  weight  and 
/  is  the  height  of  fall. 

Type  2.  Double-acting.  A  lighter  hammer-head  is  lifted  by 
steam  acting  under  its  attached  piston,  and  during  its  fall  steam 
is  admitted  above  the  piston  to  help  gravity  to  force  it  downward. 
The  energy  of  the  blow  =  Wl  (as  before)  plus  the  energy  received 
from  the  expansion  of  the  steam;  or,  if  the  steam  acts  throughout 
the  entire  stroke,  the  energy  of  b\ow  =  Wl  +  pAl,  where  p  is  the 
mean  pressure  per  square  inch  and  A  is  the  area  of  the  upper 
side  of  the  piston. 

229.  Stresses  in  Single-acting  Frames. — In  type  i,  when  the 
action  is  as  described,  a  force  acts  downward  upon  the  frame  during 
the  lifting  of  the  hammer.     The  intensity  of  this  force  =  pA  =  the 
mean  pressure  of  steam  admitted  multiplied  by  area  of  piston, 
and  the  line  of  action  is  the  axis  of  the  piston-rod.     During  the 
fall  of  the  hammer  the  cylinder  and  frame  act  simply  as  a  guide, 
and  no  force  is  applied  to  the  frame  except  such  as  may  result 
from  frictional  resistance.     The  hammer  strikes  an  anvil  which 
is  not  attached  to  the  frame,  but  rests  upon  a  separate  foundation.' 

But  a  greater  force  than  pA  may  be  applied  to  the  frame. 
In  order  that  a  cushioned  blow  may  be  struck,  the  design  is  such 
that  steam  may  be.  introduced  under  the  piston  at  any  time  during 
its  downward  movement,  and  this  steam  is  compressed  by  the 


352 


MACHINE   DESIGN. 


advancing  piston.  A  part  of  the  energy  of  the  falling  hammer 
is  used  for  this  compression.  The  pressure  in  the  cylinder  result- 
ing from  this  compression  is  communicated  to  the  lower  cylinder- 
head  and  through  it  to  the  frame.  Under  certain  conditions  steam 
might  be  admitted  at  such  a  point  of  the  stroke  that  all  of  the 
energy  of  the  falling  hammer  might  be  used  in  compressing  the 
steam  to  the  end  of  the  stroke.  The  hammer  would  then  just 
reach  the  anvil,  but  would  not  strike  a  blow. 

Fig.  264,  a  shows  by  diagram  a  hammer  of  type  i.     Steam  is 
admitted,  the  piston  is  raised,  the  exhaust-valve  is  opened,  and 


FIG.  264. 

the  piston  falls.  But  at  some  point  in  the  stroke  steam  is  again 
admitted,  filling  the  cylinder,  and  the  valve  is  closed.  Com- 
pression occurs  and  absorbs  all  or  part  of  the  energy  Wl.  In  the 
latter  case  the  hammer  will  strike  the  anvil  a  blow  whose  energy 
is  equal  to  Wl  minus  the  work  of  compressing  the  steam  in  C. 


MACHINE  FRAMES.  353 

The  compression  is  shown  upon  a  pressure-volume  diagram, 
Fig.  264,  b.  Progress  along  the  vertical  axis  from  B  toward  E 
corresponds  to  the  downward  movement  of  the  hammer.  Vertical 
ordinates  therefore  represent  space,  5,  moved  through  by  the 
hammer;  or,  since  SA  =  volume  displaced  by  the  piston,  the 
vertical  ordinates  may  also  represent  volumes.  EF  represents 

the  volume  ¥2  of  the  clearance  space,  or  —r,  the  piston  movement 

A 

which  corresponds  to  the  clearance.  Horizontal  ordinates  meas- 
ured from  BF  represent  absolute  pressures  per  square  inch. 
Let  pi  represent  the  absolute  boiler  pressure  represented  by  DK. 
TN  is  the  line  of  atmospheric  pressure.  During  the  lifting  of  the 
hammer  the  upper  surface  of  the  piston  is  exposed  to  atmospheric 
pressure  and  the  lower  surface  is  exposed  to  pressure  just  suffi- 

W 

cient  to  raise  the  hammer,  =-:-.     The  work  of  lifting  is  represented 
A. 

by  the  area  NTHG.  This  work  equals  the  energy,  Wl,  which 
the  hammer  must  give  out  in  some  way  before  reaching  the  anvil 
again.  When  the  piston  has  fallen  to  some  point,  as  D,  steam 
may  be  let  in  below  it  at  boiler  pressure,  DK.  The  advancing 
piston  will  compress  this  steam,  and  KM  will  be  the  compression 
curve.*  The  work  of  compression  is  represented  by  the  area 
RKMN.  If  the  compression  is  to  absorb  all  the  energy  Wl,  the 
area  which  represents  the  work  of  compression  must  equal  the 
area  which  represents  WL  Hence  area  NTHG  must  equal 
RKMN:  or,  since  the  area  RLGN  is  common  to  both,  the  area 
RTHL  must  equal  the  area  LKMG.  The  point  at  which  com- 
pression must  begin  in  order  to  cause  this  equality  may  be  found 
by  trial.  The  greatest  unit  pressure  reached  by  compression  is 
represented  by  EM.  The  greatest  pressure,  p-2,  upon  the  lower 
cylinder-head  is  represented  by  ATM,  since  atmospheric  pressure 

*  Assuming  pV  =consiant. 


354  MACHINE  DESIGN. 

acts  on  the  outside.  The  corresponding  total  force  communicated 
to  the  frame  =  p2A  =  P* 

If  compression  had  begun  earlier  the  energy  would  have  been 
absorbed  before  the  hammer  reached  the  anvil,  the  piston  would 
have  stopped  short  of  the  end  of  the  stroke,  the  compression  curve 
would  have  been  incomplete,  and  the  greatest  pressure  would  have 
been  less  than  EM.  Obviously  if  compression  had  begun  later 
the  greatest  pressure  would  have  been  less  than  EM.  Therefore 
the  force  P,  =p%A,  with  the  cylinder's  axis  for  its  line  of  action, 
is  the  greatest  force  that  can  be  applied  to  the  frame  in  the  regular 
working  of  the  hammer. 

A  greater  force  might  be  accidentally  applied.  For,  suppose 
that  water  is  introduced  into  the  cylinder  in  such  quantity  that 
the  piston  reaches  it  before  the  hammer  reaches  the  anvil,  then 
all  the  energy  will  be  given  out  to  overcome  the  resistance  of  the 
water.  The  resulting  force  is  indeterminate,  because  the  space 
through  which  the  resistance  acts  is  unknown.  This  force  may 
be  very  great.  The  force  applied  to  the  frame  may  be  limited 
by  the  use  of  a  "breaking- piece."  Thus  the  studs  which  hold 
on  the  lower  cylinder-head  may  be  drilled  f  so  that  they  will 
break  under  a  force  KP,  in  which  K  is  a  factor  ot  safety  and  P 
is  the  force  found  above.  Then  the  breaking-piece  will  be  safe 
under  the  maximum  working  force,  but  will  yield  when  an  acci- 
dental force  equals  KP,  thus  limiting  its  value.  The  frame  may 
be  designed  for  a  maximum  force  KP. 

230.  Problem,  Type  i.  —  Let  W,  weight  of  hammer  and 
attached  parts.  =2000  Ibs.;  I,  maximum  length  of  stroke,  =  24 
inches;  A,  effective  area  of  piston,  --=50  square  inches;  clearance 
=  15  per  cent;  boiler  pressure  =85  Ibs.  by  gauge.  Steam  is 
admitted  to  lift  the  hammer,  pressure  being  controlled  by  throttling. 

*  In  which  A  is  the   effective  area  of  the  piston,  i.e..  area  of  the  piston  less 
area  of  the  rod, 
t  See  page  136. 


MACHINE  FRAMES.  355 

The  pressure  per  square  inch  that  will  just  lift  the  hammer  = 
2000  Ibs.  H- 50  square  inches  =40  Ibs.  In  Fig.  264,  NG  repre- 
sents 40  Ibs.,  and  NT  represents  the  volume  displaced  by  the 
piston  during  a  complete  stroke.  Hence  NTHG  represents  the 
work  of  lifting  the  hammer,  or  the  energy  that  must  be  absorbed 
just  as  the  hammer  reaches  the  anvil.  Trial  shows  that  to 
accomplish  this,  compression  must  begin  at  just  about  6  inches 
from  the  end  of  the  stroke.  The  maximum  resulting  pressure, 
represented  by  NM,  equals  258  Ibs.  per  square  inch.  The  total 
pressure  acting  downward  on  the  frame  =p%A  =  258  X  50  =  12,900 
Ibs.  =P.  If  the  factor  of  safety,  K,  is  5,  the  strength  of  the 
breaking-piece  =  KP  =  5  X  12,900  =64,500  Ibs.  This  is  the  maxi- 
mum force,  and  hence  may  be  used  as  a  basis  of  the  frame  design. 

231.  Stresses   in    Double-acting   Frames.  —  In    type    2    the 
maximum  working  force  may  be  found  by  a  similar  method.     In 
Fig.  265,  NG  represents  the  pressure  per  square  inch  of  piston 
necessary  to  raise  the  hammer.     The  area  NTHG  represents 
the  energy  stored  in  the  hammer  by  lifting.     The  area  HSJL 
represents  the  work  done  by  steam  at  boiler  pressure  acting  on 
the  upper  piston  face  while  the  piston  descends  to  D.    At  this 
point  steam  is  exhausted  above  the  piston  and  let  in  below  it,  and 
compression  takes  place  during  the  remainder  of  the  stroke.     To 
absorb  all  the  energy  of  the  hammer  by  compression,  the  areas 
NTSJLG  and  RKMN  must  be  equal.     The  area  NRLG  is 
common  to  both;    hence  the  area  LKMG  must  equal  the  area 
RTSJ.    The  point  at  which  compression  must  begin  in  order 
to  cause  this  equality  may  be  found  by  trial. 

232.  Problem,  Type   2.  —  Let    W,    weight   of   hammer   and 
attached    parts,    =600  Ibs.;   /,  maximum  length  of  stroke,  =24 
inches;    A,    effective   area   of   piston  (both   faces),   =50  square 
inches;  clearance  =  15  per  cent;  boiler  pressure  =  85  Ibs.  by  gauge. 
The  construction  in  Fig.  265  shows  that  compression,  beginning 
at  9^  inches  before  the  end  of  the  piston's  stroke,  absorbs  all 


356 


MACHINE  DESIGN. 


the  energy  of  the  hammer,  and  gives  325  Ibs.  as  a  maximum 
pressure  per  square  inch.  Then  the  maximum  working  force 
=325X50  =  16,250.  If  K  =  $j  the  strength  of  the  breaking-piece 
=  16,250  X  5  =  81,250  Ibs. 


FIG.  265. 

233.  Other  Stresses  in  Hammer  Frames. — An  accidental  force 
acting  upward  may  be  applied  to  the  hammer  frame.  The  boiler 
pressure  is  necessarily  greater  than  that  which  is  necessary  to 
lift  the  hammer.*  Thus  in  §  230  a  pressure  of  40  Ibs.  per  square 
inch  is  sufficient  to  lift  the  hammer,  but  the  boiler  pressure  is 
85  Ibs.  per  square  inch.  If  the  throttle-valve  were  opened  wide 
and  held  open  during  the  movement  of  the  hammer  upward,  the 
energy  stored  in  the  hammer  when  it  reaches  its  upper  position 
would  equal  the  product  of  boiler  pressure,  piston  area,  and  length 
of  stroke,  =85X50X24  =  106,000  in. -Ibs.  The  energy  necessary 

*  So  that  it  may  be  possible  to  work  the  hammer  when  the  steam-pressure  is 
lower  in  the  boiler. 


MACHINE  FRAMES.  357 

to  just  lift  the  hammer  is  40X50X24  =  48,000  in.-lbs.  The 
difference  between  these  two  amounts  of  energy,  =58,000  in.-lbs., 
will  exist  as  kinetic  energy  of  the  moving  hammer;  and  it  must 
be  absorbed  before  the  hammer  can  be  brought  to  rest  in  its 
upper  position.  The  force  which  would  result  from  stopping 
the  hammer  would  be  dependent  upon  the  space  through  which 
the  motion  of  the  hammer  is  resisted.  Springs  are  often  provided 
to  resist  the  motion  of  the  hammer  when  near  its  upper  position. 
These  springs  increase  the  space  factor  of  the  energy  to  be  given 
out  and  thereby  reduce  the  resulting  force.  An  automatic 
device  for  closing  the  throttle -valve  before  the  end  of  the  stroke 
and  introducing  steam  for  compression  above  the  piston  may 
be  used.  The  steam  is  then  a  fluid  spring. 

234.  Design  of  Crane  Frames. — A  crane  frame  is  to  be  de- 
signed from  the  following  specifications :  Maximum  load,  5  tons 
=  10,000  Ibs.;  radius  =  maximum  distance  from  the  line  of  lifting 
to  the  axis  of  the  mast,  =  18  feet;  height  of  mast  =  20  feet;  radial 
travel  of  hook  in  its  highest  position  =  5  feet;  axis  of  jib  to  be  15 
feet  above  floor  line.  Fig.  266  shows  the  crane  indicated  by  the 
center  lines  of  its  members. 

The  external  forces  acting  on  the  crane  may  be  considered  first. 
A  load  of  10,000  Ibs.  acts  downward  in  the  line  ab.  This  is  held 
in  equilibrium  by  three  reactions :  one  acting  horizontally  toward 
the  left  through  the  upper  support,  i.e.,  along  the  line  be;  another 
acting  horizontally  toward  the  right  through  the  lower  support, 
i.e.,  in  the  line  ad;  a  third  acting  vertically  upward  at  the  lower 
support,  i.e.,  in  the  line  cd.  The  crane  is  a  "four-force  piece." 
One  force,  AB,  is  completely  known,  the  other  three  are  known 
only  in  line  of  action.  Produce  ab  and  be  to  their  intersection  at 
M.  The  line  of  action  of  the  resultant  of  ab  and  be  must  pass 
through  M.  The  resultant  of  cd  and  da  must  be  equal  and 
opposite  to  the  resultant  of  ab  and  be,  and  must  have  the  same 
line  of  action.  But  the  line  of  action  of  the  resultant  of  cd  and 


3*8 


MACHINE   DESIGN. 


da  must  pass  through  N.  Hence  MN  is  the  common  line  of 
action  of  the  resultants  of  ab  and  be  and  of  cd  and  da.  Draw 
the  vertical  line  AB  *  representing  10,000  Ibs.  upon  some  assumed 
scale;  from  B  draw  EC  parallel  to  be,  and  from  A  draw  AC  parallel 
to  MN.  The  intersection  of  these  two  lines  locates  C  and  deter- 
mines the  magnitude  of  EC.  Now  AC  is  the  resultant  of  AB 
and  BC,  and  CA,  equal  and  opposite,  is  the  resultant  of  CD 


Force  Diagram 


FIG.  266. 

and  DA.  Therefore  CA  has  but  to  be  resolved  into  vertical  and 
horizontal  components  to  determine  the  magnitudes  of  CD  and 
The  force  polygon  is  therefore  a  rectangle  and  CD=AB, 


DA. 


From  the  forces  AD  and  CD  acting  at  N,  the  supporting 
journal  and  bearing  at  the  base  of  the  crane  may  be  designed; 
and  from  the  force  BC,  acting  at  V,  the  upper  journal  and  bearing 
may  be  designed. 

235.  Jib.  —  The  forces  acting  on  the  jib  are,  first,  AB  acting 
vertically  downward  at  its  end;  second,  an  upward  reaction 

*  See  Force  Diagram,  Fig.  266. 


MACHINE  FRAMES. 


359 


from  the  brace  at  H,  whose  line  of  action  coincides  with  the  axis 
of  the  brace;*  third,  a  downward  reaction  at  L  where  the  jib 
joins  the  mast,  whose  line  of  action  must  coincide  with  the  line 
of  action  of  the  resultant  of  AB  and  the  brace  reaction.  LK  is 
therefore  this  line  of  action. 


FIG.  267. 


alb 
— JQ 


/£_ 


IS 


I' 
FIG.  268, 


In  the  force  diagram  draw  BE  parallel  to  the  center  line  of  the 
brace,  and  draw  EA  parallel  to  LK.  Then  BE  will  represent  the 
brace  reaction,  and  EA  will  represent  the  reaction  at  L  in  the  line 
ae.  Let  RQ,  Fig.  267,  represent  the  jib  isolated,  ae,  be,  and  ab 
are  the  lines  of  action  of  the  three  forces  acting  upon  it.  The 
vertical  components  of  these  forces  are  in  equilibrium,  and  tend 
to  produce  flexure  in  the  jib.  The  horizontal  components  are 
in  equilibrium  and  tend  to  produce  tension  in  the  jib.  The  vertical 


*  Considering  the  joint  between  the  brace  and  jib  equivalent  to  a  pin  con- 
nection. 


36o  MACHINE  DESIGN. 

force  acting  at  R  is  FA*  =5000  Ibs.,  the  vertical  force  acting  at 
T  is  FB,  =15,000  Ibs.,  and  the  vertical  force  acting  at  Q  is  AB, 
=  10,000  Ibs. 

Flexure  is  also  produced  in  the  jib  by  its  own  weight  acting  as 
a  uniformly  distributed  load. 

In  order  to  design  the  jib,  standard  rolled  forms  may  be 
selected  which  will  afford  convenient  support  for  the  sheave  car- 
riage. Two  channels  located  as  shown  in  Fig.  268  will  serve  for 
this  crane.  For  trial  1 2-inch  heavy  channels  are  chosen.  From 
Carnegie's  Hand-book,  the  moment  of  inertia  for  each  channel 
about  an  axis  perpendicular  to  the  web  at  the  center  =  7  =  248; 
c  =  6  inches;  the  weight,  w,  of  two  channels  per  inch  of  length 
=  8$  Ibs. 

The  total  weight  of  the  two  channels  =w/  =  8JXi8X  12  = 
1800  Ibs.  The  vertical  reaction,  PI,  at  R,  Fig.  267,  due  to  this 

weight  is  PI  =  ( —  —  )-=-^2,  from  the  equation  of  moments, 

due  to  the  weight  of  jib  about  the  point  T.  Introducing  numerical 
values,  PI  =450  Ibs.  The  total  reaction  at  R  is  therefore  5000 
—450  =  4550  Ibs.  A  diagram  of  moments  of  flexure  may  now  be 
drawn  under  the  jib,  Fig.  267.  Considering  the  portion  TQ,  the 

wli2 
moment    at    T  =  P/i+ =741,600    in.-lbs.     Divide    TQ    into 

three  equal  parts.     At  the  division  nearest    T  the  moment  = 

U'l32 

P/3+ ;  in  which /3=  the  distance  from  Q  to  the  section  con- 
sidered. The  moment  at  the  other  two  points  may  be  found 
by  similar  method. 

The  moment  at  any  point  at  the  left  of  T  =  455oX/4H — ; 

in  which  /4  is  the  distance  from  R  to  the  section  considered.  From 
the  values  thus  found  the  diagram  of  flexure  moments  may  be 

*  See  force  diagram,  Fig.  266. 


MACHINE  FRAMES. 


361 

741,600  in.-lbs. 
Ibs.    The 


drawn.    The  maximum  value  is  at  T,  where  M 

.    Me     741600X6 
The  resulting  fiber  stress  =/——««  — — = 

1  240  X  2 

horizontal  component  acting  at  R  is  equal  to  FE  (see  force  dia- 
gram, Fig.  266)  =12,000  Ibs.  An  equal  and  opposite  horizontal 
force  must  act  at  T.  Between  T  and  Q  there  is  no  tensile  stress 
due  to  the  forces  AB,  BE,  and  AE. 

FIG.  269. 


Another  force  which  modifies  this  result  needs  to  be  considered. 
Let  AB,  Fig.  269,  be  the  upper  surface  of  the  jib.  The  load  is 
supported  as  shown.  The  chain  which  is  fastened  at  B  passes 
over  the  right-hand  carriage  sheave,  down  and  under  the  hook 
sheave,  up  and  over  the  left-hand  carriage  sheave,  horizontally 
to  the  sheave  at  A,  and  thence  to  the  winding  drum.  If  a  load 
of  10,000  Ibs.  is  supported  by  the  hook  there  will  be,  neglecting 
friction,  a  tension  of  5000  Ibs.  throughout  the  entire  length  of  the 


362  MACHINE  DESIGN. 

chain  from  B  to  the  winding  drum.  There  is  therefore  a  force 
of  5000  Ibs.  tending  to  bring  A  and  B  nearer  together,  and  hence 
to  produce  compression  in  the  jib.*  The  resultant  tension  be- 
tween R  and  T  is  12,0x30—5000  =  7000  Ibs.,  while  between  T  and 
Q  there  is  a  compression  of  5000  Ibs.  The  cross-sectional  area 
of  the  two  channels  selected  =30  square  inches.  Hence  the  unit 
tensile  stress  =  /i  =  7000-^-30  =  233  Ibs.  The  maximum  unit  ten- 
sile stress  in  the  jib  =/  +  /i  ==8971  +233  =9204  Ibs.  per  square  inch. 

If  the  channels  are  of  steel,  their  unit  tensile  strength  will 
probably  equal  60,000  Ibs.  per  square  inch.  The  factor  of 
safety  =  60,000-^-9204=6.5.  In  a  crane  a  load  may  drop  through 
a  certain  space  by  reason  of  the  slipping  of  a  link  that  has  been 
caught  up,  or  the  failure  of  the  support  under  the  load  while  the 
chain  is  slack.  When  this  occurs  a  blow  is  sustained  by  the  stress 
members  of  the  crane.  The  energy  of  this  blow  equals  the  load 
multiplied  by  the  height  of  fall.  But  the  stress  members  of  the 
crane  are  long,  and  the  yielding  is  large.  Hence  the  space  through 
which  the  blow  is  resisted  is  large  and  the  resulting  force  is  less 
than  with  small  yielding.  In  other  words,  the  stress  members  act 
as  a  spring,  reducing  the  force  due  to  shock.  Hence,  in  a  crane 
of  this  type,  the  ductile  and  resilient  material  is  liable  to  modified 
shock,  and  a  factor  of  safety  =6. 5  is  large  enough. 

The  jib  might  also  be  checked  for  shear,  but  in  general  it  will 
be  found  to  have  large  excess  of  strength. 

236.  Mast. — Fig.  270  shows  the  mast  by  its  center  line  with 
the  lines  of  action  of  the  forces  acting  upon  it.  It  is  equivalent 
to  a  beam  supported  at  C  and  D  with  a  load  at  A.  The  moment 
of  flexure  at  A  equals  the  force  acting  in  the  line  fcf  multiplied 
by  the  distance  CA  in  inches,  =9000  Ibs.  X  60  inches  =  540,000 


*  There  is  also  flexure  due  to  this  force  multiplied  by  the  distance  from  the 
centre  line  of  the  horizontal  chain  to  the  gravity  axis  of  the  jib.  This  is  small 
and  may  be  neglected. 

t  The  force  BC  in  Fig.  266. 


MACHINE  FRAMES.  363 

in.-lbs.  =M.  The  flexure  moment  is  a  maximum  at  this  point. 
and  decreases  uniformly  toward  both  ends.  The  moment  diagram 
is  therefore  as  shown.  The  maximum  fiber  stress  due  to  this 
flexure  moment  =j=Mc  -j-7.  Selecting  two  light  12  -inch  channels, 

.     540000  X  6 

7  =  176;  c  =6  inches;  /  =  -  —  —  —  -r~  =9200  Ibs. 
2X1  70 

The  tension  in  the  mast  equals  the  vertical  component  of  the 
force  acting  in  the  line  ae*  =  $ooo  Ibs.  (Actually  reduced  to 
455°  by  the  effect  of  the  weight  of  the  jib.)  The  compression  in 
the  mast  due  to  the  tension  in  the  chain  =  5000  Ibs.  between  A 
and  the  point  of  support  of  the  winding  drum  B.  The  tension 
and  compression  therefore  neutralize  each  other,  except  below  B, 
where  the  flexure  moment  is  small.  Hence  the  maximum  unit 
stress  in  the  mast  is  9200  Ibs.  The  factor  of  safety  =  60000  -:~92OO 
=  6.5,  which  is  safe  as  before.  This  also  may  be  checked  for  shear. 

237.  Brace.  —  The  compression  stress  in  the  brace  is  19,000 
lbs.,t  and  the  length,  19  feet,  is  such  that  is  needs  to  be  treated 
as  a  "long  column."  Because  of  the  yielding  of  joints  and  of 
the  other  stress  members,  the  brace  is  intermediate  between  a 
member  with  "hinged  ends"  and  "flat  ends";  therefore  for  safety 
it  should  be  considered  as  hinged.  In  the  treatment  of  long 
columns,  the  "straight-line  formula"  will  be  used.!  This 
formula  is  of  the  form 


P  is  the  total  force  that  will  cause  incipient  buckling,  and  hence 
the  force  that  will  destroy  the  column;  A  is  the  cross-sectional 
area  of  the  column;  p  is  the  unit  stress  that  will  cause  buckling; 


*  The  force  AE  in  Fig.  266. 

t  See  force  diagram,  Fig.  266. 

|  For  discussion  of  long-column  formulae  see  "  Theory  and  Practice  of  Mod- 
ern Framed  Structures,"  by  Johnson,  Bryan,  and  Turneaure,  page  143.  Published 
by  John  Wiley  &  Sons. 


364  MACHINE  DESIGN. 

B  and  C  are  constants  derived  from  experiments  on  _ong  columns 
(the  values  of  B  and  C  vary  with  the  method  of  attachment  of  the 
ends  of  the  column,  and  with  the  material  of  the  column) ;  /  is 
the  length  of  the  column  in  inches;  and  r  is  the  radius  of  gyration 
of  the  cross-section,  =\/I^-A)  I  being  the  moment  of  inertia  of 
the  cross-section  referred  to  the  axis  about  which  buckling  takes 
place. 

Values  of  B  and  C  are  as  follows: 

P  I 

For  wrought  iron,  hinged  ends,  -j  =42000  —  157-. 

**•  T 

11     flat          "     ^=42000-128^. 

P  I 

'    mild  steel,       hinged     "     -j  =52500  — 220-. 

A  T 

"      "        "          flat          "     -4=52500-179-. 


The  brace  will  be  of  mild  steel  channels,  and  the  ends  will  be 
considered  as  hinged.     The  formula  to  be  used  is  therefore 


P  / 

•^  =  52500-220- 

from  which 

1  52500-220 


P  =( 


Channel  bars  may  be  selected  and  values  of  r  and  A  become 
known  from  tables.  For  trial  5-inch  light  channels  are  chosen. 
Carnegie's  tables  give  for  2  channels,  ^  =  3.8  inches,  and  A  — 
3.9  square  inches.  Introducing  these  values  in  the  above  equation, 
with  /  =  i9/Xi2=228",  gives  ^  =  153, 270  Ibs.  Since  the  maxi- 
mum compression  force  sustained  by  the  brace  =  19,000  Ibs., 


MACHINE  FRAMES. 


365 


the  factor  of  safety  =  153,270-5-19000=8  +  .  This  is  a  larger 
value  than  those  for  the  jib  and  mast,  but  it  is  probably  inadvisable 
to  use  smaller  channels  because  of  convenience  in  making  the 
connection  with  the  jib  and  mast. 

But  the  brace  must  be  made  safe  against  side  buckling.  The 
two  channels  may  be  considered  as  acting  as  a  single  member  if 
they  are  braced  laterally.  The  lateral  bracing  will  be  determined 
later.  In  Fig.  271  the  moment  of  inertia  about  the  axis  X  for 


FIG.  271. 


FIG.  272 


each  channel  =  7  (from  table.)  If  the  moment  of  inertia  of  each 
about  the  axis  Y  be  made  =  7,  the  radius  of  gyration  will  be  the 
same  about  both  axes,  the  values  in  the  above  equation  will  be 
the  same,  and  there  will  be  the  same  safety  against  side  buckling 
as  against  buckling  in  the  plane  through  the  axis  of  the  mast. 
Therefore  it  is  only  necessary  to  make  the  distance,  a,  of  each 
channel  from  the  axis  of  Y  such  that  Iy  =  7-  The  moment  of 
inertia  of  one  channel  about  its  own  gravity  axis  GG  =0.466.  Its 
moment  of  inertia  about  Y  =  7  =Ie+Ax2.  Solving,  x2  = 
+A,  whence 

7-0.466 


#-1.828. 

Hence  the  distance  apart  of  the  gravity  axes  =  1.828X2  =3.65. 
But  the  gravity  axis  is  0.44  inch  from  the  face  of  the  web,  i.e., 
x—  a  =0.44.  Therefore  the  distance,  b,  between  the  channels 
=  3.65-0.88  =  2.77  inches.  Convenience  in  construction  would 


366  MACHINE  DESIGN. 

undoubtedly  dictate  a  greater  distance,  and  hence  greater  safety 
against  side  buckling. 

The  position  of  these  two  channels  relative  to  each  other  must 
be  insured  by  some  such  means  as  diagonal  bracing.  See  Fig.  272. 
The  distance,  /,  must  be  such  that  the  channels  shall  not  buckle 
separately  under  half  the  total  load.  Solving  the  long-column 
formula  gives 


in  which  P  is  the  load  sustained  by  each  channel  (  =  19000-^-2  = 
9500  Ibs.)  multiplied  by  the  factor  of  safety,  say  6.  The  radius  of 
gyration,  r,  is  about  a  gravity  axis  parallel  to  the  web,  =\//  +A  =» 
Vo.466  -r- 1 .95  =  0.488. 

/  9500X6^0.488 

/=    52500  —  — ) =  51.8. 

\  J'95    /  22° 

The  value  of  I,  therefore,  must  not  be  greater  than  51.8  inches 
but  it  may  be  less  if  convenience,  or  the  use  of  standard  braces 
requires. 

238.  Crane  Frame  with  Tension  Rods. — The  brace  in  the 
crane  just  considered  may  be  replaced  by  tension  rods,  as  shown 
in  Fig.  273.  This  allows  the  load  to  be  moved  radially  through- 
out the  entire  length  of  the  jib.  The  force  polygon,  Fig.  274, 
shows  tension  equal  to  37,000  Ibs.  in  the  tension  rods,  and  com- 
pression in  the  jib  =  35, 800  Ibs.  If  the  tension  rods  are  made  of 
mild  steel  with  an  ultimate  tensile  strength  of  60,000  Ibs.,  and 
a  factor  of  safety  =6,  the  cross-sectional  area  must  equal 
37,000X64-60,000  =  3.7  square  inches.  If  two  rods  are  used  the 
minimum  diameter  of  each  =  1.53 5;  say,  i &  inches. 

The  mast  is  a  flexure  member  20  feet  long  supported  at  the 
ends,  and  sustaining  a  transverse  force  of  35,800  Ibs.  at  a  distance 
of  5  feet  from  the  upper  end.  The  upper  end  reaction  is  there- 


MACHINE  FRAMES. 


3^7 


fore   35,800X11  =  26,850   Ibs.,  and    the   maximum   moment   of 
flexure    at    AT  =  26,850X60  =  1,611,000    in.-lbs.     Selecting    two. 

1611000X7.5 

1 5-mch  heavy  eye-beams,  7  =  750X2;  c  =  7$;   .'. /= — 

750X2 

=  8055  Ibs.  =  maximum  unit  stress  in  the  mast.     The  factor  of 

60000 
safety  =  --  =  7.4,  safe. 


b    /     a! 

A 

I 
~~"\ 

/ 

*    \ 

13  • 

/ 

j 

i 

/a 

2 

/ 

~3T 

i 

d 

! 

/ 

35800" 

FIG.  274. 

The  moment  of  flexure  in  the  jib  is  a  maximum  when  the 
maximum  load  is  suspended  at  its  center.     The  maximum  flexure 

moment  due  to  the  load  and  the  weight  of  the  jib=M= — +-5-, 

4       o 

in  which  P=  load  =  10,000  Ibs.;  w=  weight  of  two  channels  per 
foot  of  length,  =100  Ibs.  if  12-inch  heavy  channels  are  chosen. 
Substituting  numerical  values,  M  =49,050  ft.-lbs.  =  588,600,  in.-lbs. 

The  resulting  maximum   fiber  stress   due  to  flexure,   }  =  ~J~  = 
588600X6 
- 


368  MACHINE  DESIGN- 

Compression  in  jib  due  to  chain  tension  =  10,000  Ibs.  ; 
Compression  in  jib  due  to  load  =35,800  Ibs.; 
Combined  compression  due  to  both  =  10,000X35,800  =45,800  Ibs. 
Unit  compression  =  combined  compression  -:-area  of  the  two 


channels  --—   =  1526  Ibs. 

Maximum  fiber  stress  due  to  combined  flexure  and  com- 
pression =7120  +15  26  =8647  Ibs.  The  factor  of  safety  => 

=  6.96.     If  a  smaller  factor  of  safety  were  desired,  smaller  channels 
could  be  used.    The  jib  may  be  checked  for  shear. 

The  load  might  be  moved  nearly  up  to  the  mast,  hence  the 
joint  at  F  must  be  designed  for  a  total  shear  of  10,000  Ibs.  The 
pin  and  bearing  at  G,  as  well  as  the  supporting  framework  for 
the  bearing  must  be  capable  of  sustaining  a  lateral  force  =  BC 
=8950  Ibs.  in  any  direction.  The  pivot  and  step  at  H  must  be 
capable  of  sustaining  the  lateral  force  AD  =  8950  Ibs.,  as  well  as 
a  vertical  downward  thrust  of  10,000  Ibs.  +  the  weight  of  the  crane. 

239.  Pillar-crane  Frame.  —  Fig.  275  shows  an  outline  of  the 
frame  of  a  pillar  crane.  HN  represents  the  floor  level;  HK 
represents  the  pillar,  which  is  extended  for  support  to  L;  KM 
represents  one  or  more  tension  rods;  MH  represents  the  brace. 
The  load  hangs  from  M  in  the  line  ab.  The  pillar  is  supported 
horizontally  at  H,  and  vertically  and  horizontally  at  L.  The 
force  polygon  shows  the  horizontal  forces  at  H  and  L  =  30,000  Ibs., 
and  the  vertical  force  at  L=  10,000  Ibs.  From  these  the  sup- 
ports may  be  designed.  These  supports  should  provide  for 
rotary  motion  of  the  crane  about  KL,  the  axis  of  the  pillar. 
The  brace  may  be  treated  as  in  the  jib  crane,  the  compressive 
force  being  21,400  Ibs.  The  tension  rods  may  be  designed  for 
the  force  15,800  Ibs. 

The  forces  sustained  by  the  pillar  are  as  follows  (see  Fig. 
276):  FE,  the  horizontal  component  of  AE,  =  i$,ooo  Ibs.,  acts 


MACHINE  FRAMES. 


369 


horizontally  toward  the  right  at  K,  and  EF,  the  horizontal 
component  of  EB,  =  15,000  Ibs.  acts  toward  the  left  at  H.  BC  = 
30,000  Ibs.  acts  toward  the  left  at  H,  and  DA  =30,000  Ibs.  acts 
toward  the  right  at  L.  CD  acts  upward  at  L,  producing  a  total 
compression  in  the  portion  LH  of  10,000  Ibs.  The  force  AF= 
5000  Ibs.  acts  to  produce  tension  between  H  and  K.  From 
these  data  the  pillar  may  be  designed  by  methods  already  given. 

FIG.  275. 


FIG.  276. 

240.  Frame  of  a  Steam  Riveter. — Let  Fig.  277  represent  a 
steam  riveter.  Both  the  frame  and  the  stake  are  acted  upon  by 
three  parallel  forces  when  a  rivet  is  being  driven.  The  lines  of 
action  of  these  forces  AB,  BC,  and  CA,  are  ab,  be,  ca.  The 
force  AB  required  to  drive  the  rivet  =  35,000  Ibs.  BC  and  CA 
may  be  found,  the  distances  EH  and  HG  being  known.  The 
moment  of  flexure  on  the  line  ca  =  35,000  Ibs.  X  74"  =  2, 590,000 
inch-pounds.  Let  the  line  H F  represent  this  moment.  The  moment 


37o 


MACHINE   DESIGN. 


in  any  horizontal  cross-section  may  be  found  from  the  diagram 
EFG.  Any  section  of  the  frame  or  stake  may  therefore  be  checked 
The  stake  needs  to  be  small  as  possible  in  order  that  small  boiler 
shells  and  large  flues  may  be  riveted.  In  order  that  it  may  be 
of  equal  strength,  with  the  cast  iron  frame,  it  is  made  of  material 
of  greater  unit  strength,  as  cast  steel. 


-3600535 


FIG.  277. 

The  two  bolts  which  hold  the  frame  and  stake  together  sus- 
tain a  force  of  107,000  Ibs.  The  force  upon  each  therefore  is 
53,500  Ibs.  If  the  unit  strength  of  the  material  is  50,000  Ibs., 
and  the  factor  of  safety  is  6,  the  area  of  cross-section  of  each  bolt 

6X535°° 

would  be=    5Qooo    =6.42  square  inches.      The  diameter  corre- 
sponding =  2.86  inches.      A  sfinch  bolt  has  a  diameter  at  the 


MACHINE  FRAMES.  37  1 

bottom  of  the  thread  =  2.  88  inches,  and  and  hence  3|-inch  bolts 
will  serve  as  far  as  strength  is  concerned.  But  the  body  of  the 
bolt  is  60  inches  long,  and  each  inch  of  this  length  will  yield 
a  certain  amount,  and  the  total  yielding  might  exceed  an 
allowable  value,  even  if  a  safe  stress  were  not  exceeded.  The 
yielding  per  inch  of  length,  or  the  unit  strain  =  unit  stress  -=- 
coefficient  of  elasticity,  or 


and  E=  28,000,000 

6440 

.'.  X=   0         —  =  .000023  inch. 
28000000 

Total  yielding  =  A  X6o  =  .  ooi  38  inch. 

This  amount  of  yielding  is  allowable  and  therefore  two  3^-inch 

bolts  will  serve. 


APPENDIX. 


THE  following  method  of  determining  the  position  of  the 
slider-crank  chain  corresponding  to  the  maximum  velocity  of  the 
slider  is  largely  due  to  Professor  L.  M.  Hoskins. 

Refer  to  Fig.  278. 


FIG.  278. 

BM  =  &  =  connecting-rod  length,  to  scale. 
OM  =  a  =  crank  length,  to  scale. 
OA=y. 


Angle  AOB  =90°. 
"     MAO  =  a. 


L  =2a  =  length  of  stroke  of  slider. 

From  our  study  of  the  velocity  diagram  of  the  slider-crank 
chain  (see  §  22)  we  know  that  the  length  y  will  represent  the 


373 


3/4  APPENDIX. 

velocity  of  the  slider  on  the  same  scale  as  the  length  a  repre- 
sents the  velocity  of  the  center  of  the  crank-pin.  The  length 
y  is  determined  by  erecting  at  O  a  perpendicular  to  the  line  of 
action  of  the  slider  and  cutting  this  perpendicular  by  the  con- 
necting-rod b,  extended  if  necessary. 

Our  problem,  then,  is  to  find  the  position  of  the  mechanism 
corresponding  to  the  maximum  value  of  y. 

Consider  the  triangle  whose  sides  are  y,  x,  and  a.  Calling 
the  angle  included  between  x  and  y,  a, 


—  2xy  cos  a;       .....     (i) 

(2) 


Clearing  (3)  of  fractions  and  transposing, 


2xy2=x*+x2b+y2x+y2b-a2x-a2b.     ...     (4) 
Differentiating, 


Transposing, 

dy 

dx 


.     (5) 


APPENDIX.  375 

dy 
For  maximum  value  of  y,  -£  =o;  hence  we  may  write  o  for 

the  left-hand  term  of  (5). 

—  a2  —  2y2', 


*2 (6) 

Adding  x2  and  subtracting  a2  from  both  sides  of  (6), 

From  (3), 


Substituting  this  value  in  (7), 


2a2)(x+b)  .....     (8) 
Substituting  in  (8)  the  value  of  y2  given  in  (6), 


=o  .......     (9) 

Dividing  (9)  by  a3  and  transposing, 
a2    x     x2    x3 


Equation   (10)  gives  us  the  relation  existing  between  a,  b, 
and  x  for  the  maximum  velocity  of  the  slider. 


376  APPENDIX. 

By  taking  a  series  of  values  of  T-  and  solving  (10)  for  the 

corresponding  values  of  r-,  Curve  A  has  been  constructed.    Ordi- 

x  a 

nates  are  T-,  abscissae  are  -^. 

For  any  given  problem  the   values  of  a  and  &  are  known. 
Solve  for  £. 

From  Curve  ^4  find  the  value  of  r-  corresponding  to  this  value 


From  the  determined  value  of  r-  and  the  known  value  of  b 

b 

the  numerical  value  of  x  is  found. 

But  equation  (6)  gives  for  the  maximum  value  of  y  the  rela- 
tion 


which  we  can  readily  solve  for  y  since  all  of  the  right-hand  terms 
are  now  known. 

AOB  being  a  right-angled  triangle, 


The  values  of  the  right-hand  member  being  known  we  can 
readily  solve  this  for  d. 

Let  m  represent  the  distance  moved  through  by  the  slider 
from  the  beginning  of  the  stroke,  then 

m=b+a—  d. 


APPENDIX.  377 

The  portion  of  the  stroke  accomplished  by  the  slider  at  the 
time  of  its  maximum  velocity  expressed  as  a  fraction  of  the 
whole  stroke,  2  a, 

m 

~2d' 


Curve  B  shows  the  relation  between  -r  and  — .     From  this 

0  2d 

curve  we  can  see  at  a  glance  for  any  given  value  of  -r  what  per 
cent  of  the  slider's  stroke  is  accomplished  when  its  position  of 
maximum  velocity  is  reached.  Abscissae  are  values  of  r-;  ordi- 

m 

nates  — . 
2a 

Af 

-  =  ratio  of  the  maximum  velocity  of  the  slider  to  the  velocity 
of  the  center  of  the  crank-pin.  Curve  C  shows  the  relation 

between  the  values  of  7-  and  — .    Abscissae  are  values  of  -r ;  ordi- 
o          a  b 

y  y 

nates  —  —  i.      Add  unity  to  the  ordinates  for  actual  values  of  — . 
a  a 

To  find  the  values  of  /?  corresponding  to  the  maximum  velocity 
of  the  slider  we  have  the  three  sides  of  the  triangle  OMB,  namely, 
b,  a,  and  d.  Let 


Then  cos|/?  =  -W — ,  from  which  we  can  readily  get  the 

value  of  /?. 

Curve  D  is  plotted  with  values  of  /?,  in  degrees,  as  ordinates 

and  values  of  r-  as  abscissae. 


378 


APPENDIX. 


TABLE  XXXI. — ASSUMED  VALUES  OF  — ,  AND  CORRESPONDING  COMPUTED  VALUES 
p 

OF  2-  2.  —  2.   AND  a  xo  PLOT  CURVES  A,  B,  C,  AND  D. 
ob    20.    a 


X 

~b 

O.OIO 

0.015 

0.025 

0-035 

0.050 

0.075 

O.IOO 

O.200 

a 

0.1005 

0.1234 

0.1600 

0.1902 

0.2289 

0.2832 

0.3302 

0.4817 

~b 

O.IOI 

0.1243 

0.1621 

0.1936 

0.2348 

0.2944 

0-3479 

0.5367 

m 

20. 

0.4751 

0.4704 

0.4622 

0.4561 

0.4484 

o  .  4402 

o  .  4320 

0.4239 

y. 

a 

1.005 

1.0073 

1.0131 

1.0179 

1.0258 

1.0395 

1-0533 

I.II39 

ft 

89°  55' 

89°  49' 

89°  44' 

89°  35' 

89°  26' 

88°  50' 

88°  1  6' 

85°  17' 

X 

~b 



0.300 

0.400 

0.500 

o.  600 

o.  700 

0.800 

0.900 

a 

0.6025 

0.7043 

0.7906 

0.8626 

0.9203 

0.9633 

0.9905 

L 

0.7121 

0.8854 

1.0607 

1-2393 

1.4223 

i.  6100 

1.8075 

m 

0-4273 

0.4409 

0-4597 

0.4901 

0-5374 

0.6044 

0.7221 

y. 

a 

1.1819 

1.2571 

1.3416 

1.4367 

1-5455 

1.6713 

1.8198 

f_ 



81°  23' 

76°  41' 

7i°  33' 

65°  19' 

57°  5o' 

48°  21' 

35°  8' 

APPENDIX. 


379 


INDEX. 


PAGE 

Acceleration,  diagrams  of 60-67 

Addendum  277 

Angle  of  action 284-286,  287 

Application,  machinery  of 4 

Arc  of  action 284-286,  287 

Axle  design   170-173 

Axles,   shafts,  and   spindles    170-180 

Backlash 277 

Ball-bearings,  see  Roller-  and  Ball-bearings. 

Bearing  pressure,  allowable  for  journals 182- 183 

Allowable  for  roller-  and  ball-bearings 218-220 

Allowable  for  sliding  surface  167-168 

Allowable  for  thrust-bearings  195 

Bearings 198-204 

See  also  Journals ;  Roller-  and  Ball-bearings. 

Belts  230-255 

Cone  pulley 236-239 

Crowning  pulleys 236 

Design,  theory  of  239-245 

Distance  between  shafts  for  251 

Driving  capacity,  variation  of  248-251 

Dynamo-belt  design  246-248 

Intersecting  axles  234-235 

Pump-belt  design  245-246 

Rope-drives  251-255 

Shifting,  principle  of  233 

Transmission  of  motion  by  230-233 

Twist  233-234 

Weight  of  leather  244 

Bevel-gears    304-310 


382  INDEX. 

PAGE 

Bolts  and  screws  1 19-147 

Analysis  of  screw  action  122-125 

Calculation  of  bolts  subject  to  elongation  132-136 

Calculation  of  screws  for  transmission  of  power 140-147 

Calculation  of  screws  not  stressed  in  screwing  up  125-126 

Calculation  of  screws  stressed  in  screwing  up  126-136 

Classification  and  definitions  119-132 

Design  of  bolts  for  shock  136-139 

Jam-nuts  139 

U.  S.  standard  threads  120-121 

Wrench-pull  131 

Boxes  108-204 

See  also  Journals. 

Box  pillar,  see  Supports. 

Brackets 84-85 

See  also  Supports. 


Cams 47-53 

Cast-iron  parts 85-86 

Centre    10-12 

Location  of 15-16,  17,  20-22 

Centrode  12-13 

Centres  of  three  links   16-17 

Clearance   277 

Clutches  225-229 

See  also  Couplings. 

Connecting-rod,  angularity  of 28-29 

Constrained  motion  4-8 

Cotters   156-157 

Couplings  and  clutches  221-229 

Claw  or  toothed  225-226 

Combination  friction  and  claw 229 

Compression 222-223 

Flange  222 

Flexible   224-225 

Friction    226-229 

Hooke's    224 

Oldham's    224 

Sellar's    223 

Sleeve  or  muff 221 

Weston  friction 228-229 

Cranes,  problems  in  design  of  357-369 


INDEX.  383 

PAGE 

Crank-pin,  design  of IQJ 

Cross-head  pin,  design  of  ig,2 

Cutting  speeds  30 

Cycloidal  gears 281-284,  288,  290,  291,  301,  305 

Dynamo,  belt  design  for  246-248 

Efficiency  of  machines  2 

Elements,  pairs  of  motion  x^ 

Size  of  I7_!8 

Energy,  definition  of  i 

Iti  machines   54-7O 

Law  of  conservation  of I 

Sources  of 3 

Feathers,  see  Keys. 

Fly-wheels  256-273 

Construction  of  269-273 

Design,  general  method 257 

Pump  261-264 

Punching-machine    257-261 

Steam-engine    264 

Stresses  in  arms    267-269 

Stresses  in  rims 264-267 

Theory  of  256-257 

Force,  definition  of   I 

Force-fitfis  157-161 

Frames 81-84,  337-371 

Closed 348-350 

Cranes    357-369 

Open-side   337-348 

Punching-machine    337-343 

Riveting-machine    369-371 

Slotting-machine   343-345 

Steam-engine,   center-crank    348-350 

Steam-engine,  side-crank  345-348 

Steam-hammer    35T-357 

See  also  supports  and  Machine  parts. 

Friction,  co-efficient  for  belts 244 

Co-efficient  for  dry  surfaces  228 

Co-efficinet  for  ropes 252 

Clutches  226-229 

Tower's  experiments  226-229 

See  also  Journals;  Lubrication,  and  Sliding  Surfaces. 


384  INDEX. 

PAGE 

Gearing,  see  Toothed  Wheels. 

Graphite,  as  lubricant  209 

Indicator  mechanisms  42-46 

Crosby    43-44 

Tabor    42 

Thompson    43 

Involute  gears  286-288,  289,  290,  293,  301,  306 

Instantaneous  center  10-12 

Instantaneous  motion 10 

Jib-crane 357-368 

Journals  181-209 

Allowable  bearing  pressure   182-183 

Calculation    of,    for    strength    187-193 

Crank-pin  of  engine 191 

Cross-head  pin   of   engine    192 

General  discussion  of  181-182 

Heating  of   184-186 

Lubrication  of 195-196,  201-209 

Main-journal  of  engine   188-190 

Materials  for,  and  bearings   186-187 

Proportions  of  186 

Thrust  193-197 

Tower's   experiments  on  friction  of    202-205 

See  also  Roller-  and  Ball-bearings. 

Keys  148-157 

Classification  of  148 

Cotters   156-157 

Feathers  154-155 

Kernoul  and  Barbour  152-153 

Parallel 148 

Roller   ratchet    153 

Round  taper  155 

Saddle,  flat,  and  angle  152 

Splines   154-155 

Strength  of  154 

Taper   148-149 

Woodruff 151-152 

Lathe,  bed  84 

Supports    333-336 


INDEX  385 


Legs,  see  Supports. 

Lever-crank   chain,   location   of  centres    17 

Velocity,  diagram  of  26 

Linear  velocity 22 

Points   in   different    links    26 

Line-shafts    177-180 

Linkage 14 

.Lubrication,  of  roller-  and  ball-bearings  220 

Of  rotating  surfaces 195-196,  201-209 

Of  sliding  surfaces   166-169 

Machine  cycle 2 

Machine  frames,  form  dictated  by  stress  81-84 

See  also    Frames. 

Machine  funtion  defined    2 

Machine  parts,    forms   of  cast  members    85-86 

Proportions   of    71-86 

Main  journal,  design  of   188-190 

Mechanism,  definition  of   15 

Location  of  centres  in  compound   19-22 

Motion,  chains   14-15 

Constrained    4-8 

Definition  of  r 

Free   4 

Helical   9 

Instantaneous  10-13 

Kinds  of,  in  machines  8-9 

Plane  8 

Relative    9 

Spheric    9 

Pairs  of  elements  13 

Parallel  motions,  see  Straight-line  motions. 

Passive  resistance 5-8 

Pillar  crane 368-369 

Pitch  arc 284 

Pitch  circle  276 

Pitch,  circular 277 

Diametral 277 

Planing-machine,  bed 83-84,  332,  336 

Lubrication  of   169 

Table 82-83 


386  INDEX 

PAGE 

Prime  mover   3 

Pulleys,  cone  236-239 

Crowning  236 

Idler  or  guide  235 

Proper  size  of  251 

Stresses  in  arms 267-269 

Stresses  in  rims  264-267 

See  also  Fly-wheels. 

Pump,  belt  design  for 245-246 

Fly-wheel,  design  for  261-264 

Punching-machine,  fly-wheel  design  for 257-261 

Frame,  design  for  337-343 

Quick-return  mechanisms   30-38 

Lever-crank   quick-return    32-35 

Slider-crank  quick-return 30-32 

Whitworth  quick-return  35-38 

Rigid  body 10 

Riveted  joints  87-118 

Boiler-shell  problem  1 12-1 18 

Construction  for  tightness  108-109 

Countersunk  rivets  107 

Dimensions  of  rivet-heads  106 

Efficiencies  of  various  kinds  of  94-102,  103-104 

Failure  of  91-92 

General  formulae  101-102 

Kinds  of  90-91 

Length  of  rivet 106-107 

Margin  103 

Methods  of  riveting  87-88 

More  than  two  plates  no-ill 

Nickel-steel  rivets  107 

Perforation  of  plate  88-90 

Plates  not  in  same  plane  111-112 

Plates  with  upset  edges  no 

Slippage  of  104-106 

Strength  of  materials  used  in  92-94,  109 

Strength,  proportions  and  efficiency  of  94-104 

Riveting-machine,  action  of  87 

Frame  design  for  3°9-37l 


INDEX.  387 

PAGE 

Roller-  and  ball-bearings 210-220 

Allowable  loading 218-220 

Forms  of  races  210-218 

Lubrication  and  sealing  220 

Rolling,  sliding,  and  spinning  210-217 

Size  of  219 

Rope  transmission  251-255 

Screws,  see  Bolts  and  screws. 

Screw-threads,  see  Bolts  and  screws. 

Set-screws    120,  157 

Shafting,  angular  distortion  of  176 

Combined  thrust  and  torsion  176-177 

Combined  torsion  and  bending  174-175 

Hollow  vs.  solid 175-176 

Line-shafts 177-180 

Simple  torsion  173-174 

Shaping-machine,  force  problem  56-58 

Quick-return  mechanism  for  35 

Sheave-wheels,  angle  of  groove  254 

Diameter  of 252 

Shrink-fits  157-160 

Skew  bevel-gears    311 

Slider-crank  chain,  acceleration  diagrams  60-67 

Description  of  19 

Force  problem,  shaping-machine  56-58 

Force  problems,  steam-engine 58-70 

Location  of  centres  15-16 

Maximum  velocity  of  slider  56,  373-3/9 

Tangential  effort  diagrams 68-70 

Velocity  diagram  24-26 

Sliding  surfaces 162-169 

Allowable  bearing  pressure 167- 168 

Form  of  guides  164- 166 

General  discussion  162 

Lubrication  of  166- 169 

Proportions  dictated  by  wear  163-164 

Slotted  cross-head   18-19 

Slotting-machine,  frame  design  for 343-345 

Spindles,  see  Axles,  shafts  and  spindles. 

Spiral  gears   3"-3i9 

Splines,  see  Keys. 


388  INDEX. 

PAGE 

Springs 327-331 

Cantilever  328-330 

Coil  or  spiral 330-331 

Leaf 329-331 

Spur-gears   275-304,  319-326 

Steam-engine,  boxes 199 

Crank-pin  183,  186,  191 

Cross-head  pin 183,  186,  192 

Fly-wheel  design  264 

Force  problems  58-70 

Frame,  center-crank  348-350 

Frame,  "girder  bed" 345-347 

Frame  "heavy  duty" 347-348 

Main  journal  183,  186,  188-190 

Steam-hammer,  double-acting,  frame  design  355-357 

Single-acting,  frame  design  351-355,  356 

Straight-line  motions  39-46 

General  methods  of  design  41-46 

Grasshopper  41-42 

Parallelogram  39-41 

Watt  parallel  motion  39 

Stresses  in  machine  parts,  compression 78-79 

Constant  72 

Flexure 79-80 

Shock  77-"8 

Tension 78 

Torsion  ; So-Si 

Variable  73-77 

Supports 332-336 

Divided  333-339 

General  laws  for  design  of  332 

Reduced  number  of  335-336 

Three-point  334-335 

See  also  Brackets,  and  Frames. 

Thrust  journals 193-197 

Toothed-wheels  or  gears  275-326 

Addendum  277 

Angle  of  action 284-286 

Annubar  290 

Arc  of  action 284-286 

Backlash 277 


INDEX  389 

PAGE 

Bevel-gears  3<H_3II 

Circular  pitch 2-,~ 

Clearance  277 

Cycloidal  teeth  28i,-284,  288,  290,  291,  301,  305 

Depth,  total  277 

Depth,  working  2- , 

Diametral  pitch   277 

Epicycle   trains    322-326 

Forms  of  teeth  278-284,  286-288,  294-297 

'  Interchangeable  sets 291-294 

Involute  teeth 286-288,  289,  290,  293,  301,  306 

Toothed  wheels  or  gears. 

Line  of  pressure   284 

Non-circular  wheels    303-304 

Pinion  288 

Pitch  arc    289 

Pitch  circle   276 

Proportions   of    ; 297-298 

Racks    288-289 

Reverted  trains    322 

Skew  bevel-gears    311 

Spiral  gears    311-310 

Spur-gear  chains,  compound 319-3-25 

Spur-wheels    275-304 

Strength  of  teeth   299-303,  309,  318 

Theory  of   275-276 

Worms  and  wheels   312-319 

Torque  diagrams 68-70 

Transmission,  machinery  of   i 

See  also  Belts,  Ropes,  Shafting,  and  Toothed  Wheels. 

Victor  quantity 23 

Velocity,  angular 23 

Linear  22 

Relative 22,  24,  26-27 

Whitworth  quick-return  mechanism   35-38 

Work,  definition  of i 

Worm-gearing  312-319 


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